Solution:
Let $n_{\max}$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\sqrt[m]{a b c}-\sqrt[n]{a b c}$ and since $a b c \leq 1$ we clearly have $E_{a, b, c}(m) \geq E_{a, b, c}(n)$ for $m \geq n$. So if $E_{a, b, c}(n) \geq \frac{5}{2}$ for some choice of $a, b, c \in[0,1]$, it must be $n_{\max} \leq n$. We use this remark to determine the upper bound $n_{\max} \leq 3$ by plugging some particular values of $a, b, c$ into the given inequality as follows:
$$\text{For }(a, b, c)=(1,1, c),\ c \in[0,1],\ \text{inequality (1) implies } \frac{c+2}{c+1}+\sqrt[n]{c} \leq \frac{5}{2} \Leftrightarrow \frac{1}{c+1}+\sqrt[n]{c} \leq \frac{3}{2}.$$
Obviously, every $x \in[0, 1]$ is written as $\sqrt[n]{c}$ for some $c \in[0, 1]$. So the last inequality is equivalent to:
$$\begin{aligned} & \frac{1}{x^{n}+1}+x \leq \frac{3}{2} \Leftrightarrow 2+2 x^{n+1}+2 x \leq 3 x^{n}+3 \Leftrightarrow 3 x^{n}+1 \geq 2 x^{n+1}+2 x \\ \Leftrightarrow & 2 x^{n}(1-x)+(1-x)+(x-1)\left(x^{n-1}+\cdots+x\right) \geq 0 \\ \Leftrightarrow & (1-x)\left[2 x^{n}+1-\left(x^{n-1}+x^{n-2}+\ldots+x\right)\right] \geq 0,\ \forall x \in[0,1] \end{aligned}$$
For $n=4$, the left hand side of the above becomes $(1-x)\left(2 x^{4}+1-x^{3}-x^{2}-x\right)=(1-x)(x-1)\left(2 x^{3}+x^{2}-1\right)=-(1-x)^{2}\left(2 x^{3}+x^{2}-1\right)$ which for $x=0.9$ is negative. Thus, $n_{\max} \leq 3$ as claimed.

Now, we shall prove that for $n=3$ inequality (1) holds for all $a, b, c \in[0,1]$, and this would mean $n_{\max}=3$. We shall use the following Lemma:

Lemma. For all $a, b, c \in[0, 1]: a+b+c \leq a b c+2$.

Proof of the Lemma: The required result comes by adding the following two inequalities side by side
$$\begin{aligned} & 0 \leq(a-1)(b-1) \Leftrightarrow a+b \leq a b+1 \Leftrightarrow a+b-a b \leq 1 \\ & 0 \leq(a b-1)(c-1) \Leftrightarrow a b+c \leq a b c+1 \end{aligned}$$
Because of the Lemma, our inequality (1) for $n=3$ will be proved if the following weaker inequality is proved for all $a, b, c \in[0,1]$:
$$\frac{a b c+2}{a b c+1}+\sqrt[3]{a b c} \leq \frac{5}{2} \Leftrightarrow \frac{1}{a b c+1}+\sqrt[3]{a b c} \leq \frac{3}{2}$$
Denoting $\sqrt[3]{a b c}=y \in[0, 1]$, this inequality becomes:
$$\begin{aligned} & \frac{1}{y^{3}+1}+y \leq \frac{3}{2} \Leftrightarrow 2+2 y^{4}+2 y \leq 3 y^{3}+3 \Leftrightarrow -2 y^{4}+3 y^{3}-2 y+1 \geq 0 \\ \Leftrightarrow & 2 y^{3}(1-y)+(y-1) y(y+1)+(1-y) \geq 0 \Leftrightarrow (1-y)\left(2 y^{3}+1-y^{2}-y\right) \geq 0 \end{aligned}$$
The last inequality is obvious because $1-y \geq 0$ and $2 y^{3}+1-y^{2}-y=y^{3}+(y-1)^{2}(y+1) \geq 0$.