Solution:

Notice that, if $p$ is a prime and $g$ is a polynomial with integer coefficients such that $g(n) \equiv 0 \pmod{p}$ for some $n$, then $g(n + m p)$ is divisible by $p$ as well for any integer multiple $m p$ of $p$. Therefore, it suffices to find the smallest possible degree of a polynomial $f$ for which $f(0), f(1), f(2), \ldots, f(16)$ are divisible by $17$ and by $5$.

There is a polynomial of degree $17$ with integer coefficients having $f(0) = f(1) = \cdots = f(16) = 0$, namely $f(x) = (x)(x-1)(x-2) \cdots (x-16)$. Thus the minimal degree is no larger than $17$.

Now, let $f$ be such a polynomial and consider $f$ modulo $17$. The polynomial has $17$ roots, so it must be at least degree $17$ when taken modulo $17$. Thus $f$ has degree at least $17$ as well.