17. Let $S=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid 0 \leqslant x_{i} \leqslant p-1, i=1,2, \cdots, n\right\}$, then $|S|=$ $p^{n}$. Consider the following sum:
$$T=\sum_{\left(y_{1}, y_{2}, \cdots, y_{n}\right) \in S} f\left(y_{1}, y_{2}, \cdots, y_{n}\right)^{p-1}$$

Notice that, the degree of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is less than $n$, hence the degree of each term in the expansion of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)^{p-1}$ is less than $(p-1) n$. Thus, when each term in $T$ is expanded and like terms are combined, the term $A y_{1}^{q_{1}} \cdot y_{2}^{q_{2}} \cdot \cdots \cdot y_{n}^{q_{n}}, \alpha_{i} \in \mathbf{N}$ satisfies:
$$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}<(p-1) n$$

Therefore, for this term, there exists an $i$ such that $\alpha_{i} \leqslant p-2$. Fixing $y_{1}, y_{2}, \cdots, y_{i-1}, y_{i+1}, \cdots, y_{n}$ and letting $y_{i}$ vary from 0 to $p-1$, using the conclusion from Example 1 in Section 3.2, we have:
$$\sum_{y_{i}=0}^{p-1} A y_{1}^{q_{1}} \cdot y_{2}^{q_{2}} \cdots \cdots y_{n}^{q_{n}} \equiv 0(\bmod p)$$

Thus, $\sum_{\left(y_{1}, y_{2}, \cdots, y_{n}\right) \in S} A y_{1}^{q_{1}} \cdot y_{2}^{q_{2}} \cdot \cdots \cdot y_{n}^{q_{n}} \equiv 0(\bmod p)$, and hence $T \equiv 0(\bmod p)$.
On the other hand, by Fermat's Little Theorem, we know:
$$f\left(y_{1}, y_{2}, \cdots, y_{n}\right)^{p-1} \equiv\left\{\begin{array}{ll} 1(\bmod p), & f\left(y_{1}, y_{2}, \cdots, y_{n}\right) \neq 0(\bmod p), \\ 0(\bmod p), & f\left(y_{1}, y_{2}, \cdots, y_{n}\right) \equiv 0(\bmod p) . \end{array}\right.$$

Thus, from $T \equiv 0(\bmod p)$, we know that the number of arrays $\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ in $S$ such that $f\left(y_{1}, y_{2}, \cdots, y_{n}\right) \neq \equiv(\bmod p)$ is a multiple of $p$. Combining this with $|S|=p^{n}$, we obtain the desired conclusion.