If $x$ is a rational multiple of $\pi$, for a natural number $N$ big enough $\sin(n!\,x) = 0$ for all $n\geqslant N$ and then $\sin(n!\,x)\to 0$ as $n\to +\infty$.
However, I'm not so sure about the converse anymore: If it holds that $\sin(n!\,x)\to 0$ as $n\to +\infty$, does it necessarily follow that $x$ belongs to $\pi\mathbf{Q}$?
Here we extend the idea discussed in the comments and discuss the condition for $x$ that makes the limit zero.
Let $x = \pi r$. Write the fractional part $\{r\} = r -\lfloor r \rfloor$ in the form $$ \{r\} = \sum_{j=2}^{\infty} \frac{a_j}{j!} \quad \text{for} \quad a_j \in \{\,0, 1,\ldots, j-1\,\}. $$
Then we note that $$ \begin{split} |\sin(n!\, \pi r)| = \Biggl| \sin\biggl(n!\, \pi\sum_{j=2}^{\infty} \frac{a_j}{j!} \biggr) \Biggr| = \Biggl| \sin\biggl(\pi \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)(n+2)\cdots(n+k)} \biggr) \Biggr| \end{split} $$ and $$ \begin{split} \frac{a_{n+1}}{n+1} &\leq \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)(n+2)\cdots (n+k)} \\ &\leq \frac{a_{n+1}}{n+1} +\sum_{k=2}^{\infty} \frac{n+k-1}{(n+1)(n+2)\cdots (n+k)} \\ &= \frac{a_{n+1} + 1}{n+1}. \end{split} $$
From this, it is not hard to check that $\sin(n!\, \pi r) \to 0$ if and only if the sequence $(a_{n+1}/(n+1))$ has only limit points in $\{\,0, 1\,\}$, or equivalently, $\operatorname{dist}(a_{n+1}/(n+1), \mathbb{Z})\to 0$. Thomas's example corresponds to the case where all $a_j$ are either $0$ or $1$.
Finally, since the set of all such $r$ is uncountable, there is an irrational number $r$ such that $\sin(n!\, \pi r) \to 0$.