How many subgroups of order 17 does $S_{17}$ have?

Your answer looks correct.

It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.

An alternative (but not necessarily better) proof is as follows:

Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.

For each subgroup $G$ fix $g\in G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.

Write $g=(1,2,x_3,\ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.