First suppose $b \equiv \pm a \pmod{p}$ and $c \equiv \pm b \pmod{p}$. Then, for any $n$, we have $a^n + b^n + c^n \equiv \pm a^n$ or $\pm 3a^n \pmod{p}$. We are given that $p \neq 3$ (since $\frac{3-1}{2}$ is not prime) and $p \nmid a$, so it follows that $a^n + b^n + c^n \not\equiv 0 \pmod{p}$. The claim is trivial in this case. Otherwise, we may assume without loss of generality that $b \not\equiv \pm a \pmod{p} \Longrightarrow b a^{-1} \not\equiv \pm 1 \pmod{p}$.

Now let $q = \frac{p-1}{2}$. By Fermat's little theorem, we know that the order of $b a^{-1} \pmod{p}$ divides $p-1 = 2q$. However, since $b a^{-1} \not\equiv \pm 1 \pmod{p}$, the order of $b a^{-1}$ does not divide 2. Thus, the order must be either $q$ or $2q$.

Next, let $S$ denote the set of positive integers $n < p$ such that $a^n + b^n + c^n \equiv 0 \pmod{p}$, and let $s_t$ denote the number of ordered pairs $(i, j) \subset S$ such that $i - j \equiv t \pmod{p-1}$.

Lemma: If $t$ is a positive integer less than $2q$ and not equal to $q$, then $s_t \leq 2$.

Proof: Consider $i, j \in S$ with $j - i \equiv t \pmod{p-1}$. Then we have

$$\begin{aligned} & a^i + b^i + c^i \equiv 0 \pmod{p} \\ \Longrightarrow & a^i c^{j-i} + b^i c^{j-i} + c^j \equiv 0 \pmod{p} \\ \Longrightarrow & a^i c^{j-i} + b^i c^{j-i} - a^j - b^j \equiv 0 \pmod{p} \\ \Longrightarrow & a^i \cdot (c^t - a^t) \equiv b^i \cdot (b^t - c^t) \pmod{p}. \end{aligned}$$

If $c^t \equiv a^t \pmod{p}$, then this implies $c^t \equiv b^t \pmod{p}$ as well, so $(a b^{-1})^t \equiv 1 \pmod{p}$. However, we know the order of $a b^{-1}$ is $q$ or $2q$, and $q \neq t$, so this is impossible. Thus, we can write

$$(a b^{-1})^i \equiv (b^t - c^t) \cdot (c^t - a^t)^{-1} \pmod{p}.$$

For a fixed $t$, the right-hand side of this equation is fixed, so $(a b^{-1})^i$ is also fixed. Since the order of $a b^{-1}$ is either $q$ or $2q$, it follows that there are at most 2 solutions for $i$, and the lemma is proven.

Now, for each element $i$ in $S$, there are at least $|S| - 2$ other elements that differ from $i$ by a quantity other than $q \pmod{p-1}$. Therefore, the lemma implies that

$$\begin{aligned} |S| \cdot (|S| - 2) & \leq \sum_{t \neq q} s_t \leq 2 \cdot (p-2) \\ \Longrightarrow \quad (|S| - 1)^2 & \leq 2p - 3 \\ \Longrightarrow \quad |S| & < \sqrt{2p} + 1. \end{aligned}$$

1. Let $p$ be a prime number such that $\frac{p-1}{2}$ is also prime. Define $q = \frac{p-1}{2}$. We are given that $a, b, c$ are integers not divisible by $p$. Without loss of generality, we can assume $c = 1$ by scaling.

2. Define the set $S \subset \mathbb{Z}/(p-1)\mathbb{Z}$ to be the set of all $n$ such that $a^n + b^n \equiv -1 \pmod{p}$.

3. We first consider the case where $\mathrm{ord}_p(a/b) = 1$. This implies $a = b$. In this case, we need $a^n \equiv -1/2 \pmod{p}$. Since $p \neq 2, 3$, $-1/2 \not\in \{-1, 1\} \pmod{p}$. Therefore, at most two values of $n$ satisfy this condition.

4. Next, consider the case where $\mathrm{ord}_p(a/b) = 2$. This implies $a = -b$. In this case, we need $n$ to be even, which yields the same equation as before. Hence, $\mathrm{ord}_p(a/b) \in \{q, 2q\}$.

5. Claim: Each difference between distinct elements of $S$ shows up at most twice, except for $q$.

6. Proof of Claim: Suppose we have $k \neq l$ (in $\mathbb{Z}/(p-1)\mathbb{Z}$) with $k, l, k+d, l+d \in S$ for some $d \neq 0$. This implies:
$$a^{l+d}b^k + a^k b^{l+d} \equiv a^{k+d}b^l + a^l b^{k+d} \pmod{p}$$
Simplifying, we get:
$$a^k b^l (a^d - b^d) \equiv a^l b^k (a^d - b^d) \pmod{p}$$
Hence, either $(a/b)^d \equiv 1 \pmod{p}$ or $(a/b)^{k-l} \equiv 1 \pmod{p}$. If $d \neq q$, then this means $k - l = q$, which implies the desired claim. $\blacksquare$

7. For any element of $S$, there are at least $|S| - 2$ other elements which don't differ from it by $q$. Hence, we need:
$$|S|(|S| - 2) \leq 2(p - 2)$$
Solving for $|S|$, we get:
$$|S|^2 - 2|S| \leq 2(p - 2)$$
$$|S|^2 - 2|S| - 2(p - 2) \leq 0$$
Solving this quadratic inequality, we find:
$$|S| \leq 1 + \sqrt{2p}$$

The final answer is $\boxed{ 1 + \sqrt{2p} }$