Here is a proof that the answer is (almost certainly) not provable using current techniques. We will prove that the series in fact diverges if $2\pi e$ is a rational number with a prime numerator. We first prove the following claims:
Lemma 1. If $p$ is an odd prime number and $S\subset \mathbb Z$ so that
$$\sum_{s\in S}e^{2\pi i s/p}\in\mathbb R,$$
then $\sum_{s\in S}s\equiv 0\bmod p$.
Proof. Let $\zeta=e^{2\pi i/p}$. We have
$$\sum_{s\in S}\zeta^s=\sum_{s\in S}\zeta^{-s},$$
since the sum is its own conjugate. As a result, since the minimal polynomial of $\zeta$ is $\frac{\zeta^p-1}{\zeta-1}$, we see
$$\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s}-x^{p-s}\right),$$
where we have placed each element of $s$ in $[0,p)$. The polynomial on the left is coprime with $x-1$ and the polynomial on the right has it as a factor, so
$$\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s-1}+\cdots+x^{p-s}\right).$$
Now, the quotient of these two polynomials must be an integer polynomial, so in particular the value of the left-side polynomial at $1$ must divide the value of the right-side polynomial at $1$. This gives $p|\sum_{s\in S}2s,$ finishing the proof.
Define
$$a_n=\sum_{k=0}^n \frac{n!}{k!}.$$
Lemma 2. If $p$ is a prime number,
$$\sum_{n=0}^{p-1}a_n\equiv -1\bmod p.$$
Proof.
\begin{align*}
\sum_{n=0}^{p-1}a_n
&=\sum_{0\leq k\leq n\leq p-1}\frac{n!}{k!}\\
&=\sum_{0\leq n-k\leq n\leq p-1}(n-k)!\binom n{n-k}\\
&=\sum_{j=0}^{p-1}\sum_{n=j}^{p-1}n(n-1)\cdots(n-j+1)\\
&\equiv \sum_{j=0}^{p-1}\sum_{n=0}^{p-1}n(n-1)\cdots(n-j+1)\pmod p,
\end{align*}
where we have set $j=n-k$. The inside sum is a sum of a polynomial over all elements of $\mathbb Z/p\mathbb Z$, and as a result it is $0$ as long as the polynomial is of degree less than $p-1$ and it is $-1$ for a monic polynomial of degree $p-1$. Since the only term for which this polynomial is of degree $p-1$ is $j=p-1$, we get the result.
Now, let $2\pi e = p/q$. Define $\mathcal E(x)=e^{2\pi i x}$ to map from $\mathbb R/\mathbb Z$, and note that $\mathcal E(x+\epsilon)=\mathcal E(x)+O(\epsilon)$. We have
\begin{align*}
\sin((n+p)!)
&=\operatorname{Im}\mathcal E\left(\frac{(n+p)!}{2\pi}\right)\\
&=\operatorname{Im}\mathcal E\left(\frac{qe(n+p)!}{p}\right).
\end{align*}
We will investigate $\frac{qe(n+p)!}{p}$ "modulo $1$." We see that
\begin{align*}
\frac{qe(n+p)!}{p}
&=q\sum_{k=0}^\infty \frac{(n+p)!}{pk!}\\
&\equiv q\sum_{k=n+1}^\infty \frac{(n+p)!}{pk!}\pmod 1\\
&=O(1/n)+q\sum_{k=n+1}^{n+p}\frac{(n+p)!}{pk!}\\
&=O(1/n)+\frac qp\left[\sum_{k=n+1}^{n+p}\frac{(n+p)!}{k!}\pmod p\right].
\end{align*}
Now,
\begin{align*}
\sum_{k=n+1}^{n+p}\frac{(n+p)!}{k!}=\sum_{j=0}^{p-1}\frac{(n+p)!}{(n+p-j)!}
&=\sum_{j=0}^{p-1}(n+p)(n+p-1)\cdots(n+p-j+1)\\
&\equiv \sum_{j=0}^{p-1}m(m-1)\cdots (m-j+1)\pmod p,
\end{align*}
where $m$ is the remainder when $n$ is divided by $p$. The terms with $j>m$ in this sum go to $0$, giving us
$$\sum_{j=0}^m \frac{m!}{(m-j)!}=a_m.$$
Putting this together, we see that
$$\sin((n+p)!)=\operatorname{Im}\mathcal E\left(\frac{qa_{n\bmod p}}p\right)+O\left(\frac 1n\right).$$
In particular, the convergence of our sum would imply, since the $O(1/n)$ terms give a convergent series when multiplied by $O(1/n)$, that
$$x_N=\operatorname{Im}\sum_{n=1}^N\frac 1n\mathcal E\left(\frac{qa_{n\bmod p}}p\right)$$
should converge. In particular, $\{x_{pN}\}$ must converge, which implies that
$$\sum_{m=0}^{p-1}\mathcal E\left(\frac{qa_m}p\right)$$
must be real (as otherwise the series diverges like the harmonic series). By Lemma 1, this implies that
$$\sum_{m=0}^{p-1}a_m=0\bmod p,$$
which contradicts Lemma 2.
Best Answer
I summarize below the main ideas of the paper [1] whose link is in my comment above:
$$ W_k\ge\frac12 k^{77/76} $$ meaning they are pretty scarce.