Consider the generating function of $\{f(n)\}$:
$$\begin{align*} G(t) &= \sum_{n=0}^{\infty} f(n) t^n \\ &= (1 + t^a + t^{2a} + \dots)(1 + t^b + t^{2b} + \dots)(1 + t^c + t^{2c} + \dots) \\ &= \frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)}. \end{align*}$$
Since $a$, $b$, $c$ are pairwise coprime, the denominator $(1-t^a)(1-t^b)(1-t^c)$ contains no repeated factors other than $(1-t)^3$. Let $\omega = e^{2\pi i/a}$, $\tau = e^{2\pi i/b}$, $\rho = e^{2\pi i/c}$ be the primitive $a$, $b$, $c$th roots of unity, respectively. Consider the partial fraction decomposition
$$\frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)} = \frac{h_0}{(1 - t)} + \frac{h_1}{(1 - t)^2} + \frac{h_2}{(1 - t)^3} \\ + \sum_{k=1}^{a-1} \frac{d_k}{1 - \omega^k t} + \sum_{k=1}^{b-1} \frac{e_k}{1 - \tau^k t} + \sum_{k=1}^{c-1} \frac{f_k}{1 - \rho^k t},$$
in which the numerators $h_0$, $h_1$, $h_2$, $d_1, \dots, d_{a-1}$, $e_1, \dots, e_{b-1}$, $f_1, \dots, f_{c-1}$ are all complex numbers. To find $d_k$ in $\frac{d_k}{1 - \omega^k t}$, multiply $(1 - \omega^k t)$ on both sides and let $t \to \omega^{-k}$ (or just let $1 - \omega^k t = 0$). As all the other terms vanish, we arrive at
$$\begin{align*} d_k &= \lim_{t \to \omega^{-k}} \frac{1 - \omega^k t}{(1 - (\omega^k t)^a)(1 - t^b)(1 - t^c)} \\ &= \frac{1}{a} \times \frac{1}{(1 - \omega^{-kb})(1 - \omega^{-kc})}. \end{align*}$$
In the expansion of $G(t)$, the coefficient of $t^n$ is
$$f(n) = h_0 + h_1 \times (n + 1) + h_2 \times \frac{(n + 1)(n + 2)}{2} \\ + \sum_{k=1}^{a-1} d_k \omega^{kn} + \sum_{k=1}^{b-1} e_k \tau^{kn} + \sum_{k=1}^{c-1} f_k \rho^{kn}.$$

Observe that
$$\begin{aligned} \left| \sum_{k=1}^{a-1} d_k \omega^{kn} \right| & \le \sum_{k=1}^{a-1} |d_k| \\ & = \frac{1}{a} \times \sum_{k=1}^{a-1} \frac{1}{|1 - \omega^{-kb}| |1 - \omega^{-kc}|} \\ & \le \frac{1}{a} \times \sum_{k=1}^{a-1} \frac{1}{|1 - \omega^k|^2} \\ & = \frac{a^2 - 1}{12a} < \frac{a}{12}. \end{aligned}$$
Here, the equality in the last row requires a lemma which will be proved.
Likewise, we have in the following $\left|\sum_{k=1}^{b-1} e_k \tau^{kn}\right| < \frac{b}{12}$ and $\left|\sum_{k=1}^{c-1} f_k \rho^{kn}\right| < \frac{c}{12}$.
Taking $\beta_2 = \frac{h_2}{2}$, $\beta_1 = h_1 + \frac{3}{2}h_2$, $\beta_0 = h_0 + h_1 + h_2$, it follows that
$$\begin{aligned} |\beta_2 n^2 + \beta_1 n + \beta_0 - f(n)| & \le \left|\sum_{k=1}^{a-1} d_k \omega^{kn}\right| + \left|\sum_{k=1}^{b-1} e_k \tau^{kn}\right| + \left|\sum_{k=1}^{c-1} f_k \rho^{kn}\right| \\ & < \frac{a+b+c}{12}. \end{aligned}$$

Lemma (for the equality in the last row) For any positive integer $m$,
$$\sum_{k=1}^{m-1} \frac{1}{\sin^2 \frac{k\pi}{m}} = \frac{m^2-1}{3} \text{ holds; equivalently, } \sum_{k=1}^{m-1} \cot^2 \frac{k\pi}{m} = \frac{m^2-3m+2}{3}.$$
Proof of lemma For $\vartheta \in A = \left\{ \frac{\pi}{m}, \frac{2\pi}{m}, \dots, \frac{(m-1)\pi}{m} \right\}$,
$$(\cos \vartheta + i \sin \vartheta)^m = \cos(m\vartheta) + i \sin(m\vartheta) = \pm 1,$$
where the imaginary part
$$C_m^1 \sin \vartheta \cos^{m-1} \vartheta - C_m^3 \sin^3 \vartheta \cos^{m-3} \vartheta + C_m^5 \sin^5 \vartheta \cos^{m-5} \vartheta + \dots = 0,$$
or $C_m^1(\cot \vartheta)^{m-1} - C_m^3(\cot \vartheta)^{m-3} + C_m^5(\cot \vartheta)^{m-5} + \dots = 0$. This implies that $\{\cot \frac{k\pi}{m} : k = 1, 2, \dots, m-1\}$ satisfy the $(m-1)$th degree polynomial in cot $\vartheta$. According to Viète's formulas, the sum of the $m-1$ roots is $\sigma_1 = 0$ while the sum of their products multiplied in pairs is
$$\sigma_2 = - \frac{C_m^3}{C_m^1} = - \frac{(m-1)(m-2)}{6}.$$
Consequently, the sum of the squares of the $m-1$ roots is
$$\sum_{k=1}^{m-1} \cot^2 \frac{k\pi}{m} = \sigma_1^2 - 2\sigma_2 = \frac{m^2 - 3m + 2}{3}. \quad \Box$$

Solution:

By symmetry, assume $a \le b \le c$. If $a = b = c = 1$, $f(n)$ is the number of ways of writing $n$ as three nonnegative integers, which is $C_{n+2}^2$, and by taking $\alpha = \frac{1}{2}$, $\beta = \frac{3}{2}$, $\gamma = 1$, the error is constantly 0. In the following, assume $c > 1$. For nonnegative integer $m$, let $g(m)$ represent the number of nonnegative integer solutions $(x, y)$ of $a x + b y = m$.
First, we try to find $g(m)$. Let $x \equiv k a \pmod b$, $m \equiv \ell b \pmod a$, where
$$k \in \{0, 1, \dots, b-1\}, \quad \ell \in \{0, 1, \dots, a-1\}.$$
Evidently, $x \equiv k \pmod b$, $y \equiv \ell \pmod a$. Suppose $x = k + u b$, $y = \ell + v a$.
Then
$$m = k a + \ell b + (u + v) a b,$$
giving $u + v = \frac{m - (k a + \ell b)}{a b}$, and the number of pairs $(u,v)$ is
$$\frac{m - (k a + \ell b)}{a b} + 1$$ (note that $\frac{m - (k a + \ell b)}{a b}$ is an integer greater than $-2$; when $\frac{m - (k a + \ell b)}{a b} = -1$ the conclusion is still valid). Therefore,
$$g(m) = \frac{m - (k a + \ell b)}{a b} + 1.$$
Next, consider $f(n)$. In the Diophantine equation $a x + b y + c z = n$, if $z$ is given ($c z \le n$), then the number of nonnegative solutions of $a x + b y = n - c z$ is $g(n - c z)$. Hence,
$$f(n) = \sum_{i=0}^{\lfloor n/c \rfloor} g(n - i c).$$
For each integer $i$, define $k_i \in \{0, 1, \dots, b-1\}$ and $\ell_i \in \{0, 1, \dots, a-1\}$ such that
$$n - i c = k_i a \pmod{b}, \quad n - i c = \ell_i b \pmod{a}.$$
We have
$$\begin{aligned} f(n) &= \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} g(n - i c) \\ &= \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} \left( \frac{(n - i c) - k_i a + \ell_i b}{a b} + 1 \right) \\ &= \lfloor \frac{n}{c} \rfloor \cdot \left( \frac{n}{a b} + 1 \right) - \frac{c}{2 a b} \lfloor \frac{n}{c} \rfloor \left( \lfloor \frac{n}{c} \rfloor + 1 \right) - \left( \frac{1}{b} \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} k_i + \frac{1}{a} \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} \ell_i \right). \end{aligned}$$
Note that $a$, $b$, $c$ are pairwise coprime. Hence, for $i = 0, 1, \dots, b-1$, the integers $n - i c$ cover a complete system of residues modulo $b$; so do the integers $k_i a$ and the integers $k_i$. In addition, when $i$ shifts by a multiple of $b$, $k_i$ does not change.
Now we turn to $\sum_{i=0}^{\lfloor n/c \rfloor} \left( k_i - \frac{b-1}{2} \right)$: notice that the summands add up to 0 as $i$ goes over a complete system of residues modulo $b$. Therefore, $\sum_{i=0}^{\lfloor n/c \rfloor} \left( k_i - \frac{b-1}{2} \right) = \sum_{i=0}^{s} \left( k_i - \frac{b-1}{2} \right)$, where $s$ is the least nonnegative residue of $\lfloor \frac{n}{c} \rfloor$ modulo $b$. In $\sum_{i=0}^{s} \left( k_i - \frac{b-1}{2} \right)$, the summands are distinct and are all from $\left\{ \frac{1-b}{2}, \frac{3-b}{2}, \dots, \frac{b-3}{2}, \frac{b-1}{2} \right\}$ (the positive and negative elements are symmetric). This indicates that $\left| \sum_{i=0}^{s} \left( k_i - \frac{b-1}{2} \right) \right|$ cannot exceed the sum of the positive elements $\frac{2}{2} + \frac{4}{2} + \dots + \frac{b-3}{2} + \frac{b-1}{2} = \frac{b^2-1}{8}$ when $b$ is odd, or $\frac{1}{2} + \frac{3}{2} + \dots + \frac{b-3}{2} + \frac{b-1}{2} = \frac{b^2}{8}$ when $b$ is even. Hence,
$$\left| \sum_{i=0}^{\lfloor n/c \rfloor} \left( k_i - \frac{b-1}{2} \right) \right| \le \frac{b^2}{8}. \qquad \textcircled{1}$$
Likewise,
$$\left| \sum_{i=0}^{\lfloor n/c \rfloor} \left( \ell_i - \frac{a-1}{2} \right) \right| \le \frac{a^2}{8}. \qquad \textcircled{2}$$
From ① and ②, it follows that
$$\begin{aligned} & \left| \left( \frac{1}{b} \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} k_i + \frac{1}{b} \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} l_i \right) - \left( \lfloor \frac{n}{c} \rfloor + 1 \right) \left( \frac{b-1}{2b} + \frac{a-1}{2a} \right) \right| \\ & \le \frac{1}{b} \left| \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} \left( k_i - \frac{b-1}{2} \right) \right| + \frac{1}{a} \left| \sum_{i=0}^{\lfloor \frac{n}{c} \rfloor} \left( \ell_i - \frac{a-1}{2} \right) \right| \\ & \le \frac{b}{8} + \frac{a}{8}. \end{aligned} \qquad \textcircled{3}$$
Let $r$ be the least nonnegative residue of $n$ modulo $c$. Then
$$\begin{aligned} & \lfloor \frac{n}{c} \rfloor \cdot \left( \frac{n}{a b} + 1 \right) - \frac{c}{2 a b} \lfloor \frac{n}{c} \rfloor \cdot \left( \lfloor \frac{n}{c} \rfloor + 1 \right) - \left( \lfloor \frac{n}{c} \rfloor + 1 \right) \left( \frac{b-1}{2b} + \frac{a-1}{2a} \right) \\ & = \frac{(n-r)(n+ab)}{abc} - \frac{(n-r)(n-r+c)}{2abc} - \frac{(2ab-a-b)(n-r+c)}{2abc} \\ & = \frac{n^2 - (a+b+c)n - (2abc+ac+bc) - r(r+a+b-c)}{2abc}. \end{aligned} \qquad \textcircled{4}$$
As $r$ takes values $0, 1, \dots, c-1$, denote the maximum and the minimum of $r(r+a+b-c)$ as $A+B$ and $A-B$, respectively. By ③ and ④,
$$\left| f(n) - \frac{n^2 - (a+b+c)n - (2abc+ac+bc) - A}{2abc} \right| \le \frac{a+b}{8} + \frac{B}{2abc}.$$
Let $\alpha = \frac{1}{2abc}$, $\beta = -\frac{a+b+c}{2abc}$,
$$\gamma = -\frac{2abc + ac + bc + A}{2abc}.$$
We will justify the problem statement
$$|f(n) - (\alpha n^2 + \beta n + \gamma)| < \frac{a+b+c}{12} \qquad \textcircled{5}$$
for different $a$, $b$, $c$ values. $\square$