Solution:
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 $\pm 3 a^{n} \pmod{p}$. 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 $b a^{-1} \bmod p$ divides $p-1=2 q$. 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 $2 q$.

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 $2 q$ 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
$$\begin{aligned} & a^{i}+b^{i}+c^{i} \equiv 0 \pmod{p} \\ \Longrightarrow & a^{i} c^{j-i}+b^{i} c^{j-i}+c^{j} \equiv 0 \pmod{p} \\ \Longrightarrow & a^{i} c^{j-i}+b^{i} c^{j-i}-a^{j}-b^{j} \equiv 0 \pmod{p} \\ \Longrightarrow & a^{i} \cdot\left(c^{t}-a^{t}\right) \equiv b^{i} \cdot\left(b^{t}-c^{t}\right) \pmod{p} . \end{aligned}$$
If $c^{t} \equiv a^{t} \pmod{p}$, then this implies $c^{t} \equiv b^{t} \pmod{p}$ as well, so $\left(a b^{-1}\right)^{t} \equiv 1 \pmod{p}$. However, we know the order of $a b^{-1}$ is $q$ or $2 q$, and $q \nmid t$, so this is impossible. Thus, we can write
$$\left(a b^{-1}\right)^{i} \equiv\left(b^{t}-c^{t}\right) \cdot\left(c^{t}-a^{t}\right)^{-1} \pmod{p}$$
For a fixed $t$, the right-hand side of this equation is fixed, so $\left(a b^{-1}\right)^{i}$ is also fixed. Since the order of $a b^{-1}$ is either $q$ or $2 q$, 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
$$\begin{aligned} |S| \cdot(|S|-2) & \leq \sum_{t \neq q} s_{t} \leq 2 \cdot(p-2) \\ \Longrightarrow \quad(|S|-1)^{2} & \leq 2 p-3 \\ \Longrightarrow \quad|S| & <\sqrt{2 p}+1 . \end{aligned}$$