This is true. It doesn't have much to do with the details of $\sin(\sin(\ ))$. Rather:

**Theorem** Let $f: \mathbb{R} \to \mathbb{R}$ be any smooth, $2 \pi$-periodic function with $\int_{z=0}^{2 \pi} f(z) dz=0$. Then $\sum_{n=1}^N f(n)$ is bounded independent of $N$.

There are two ingredients in the proof. First, we bound the coefficients of the Fourier series of $f$. We can write
$$f(\theta) = \sum_{k=-\infty}^{\infty} a_k e^{i k \theta}$$
where
$$a_k = \frac{1}{2 \pi} \int_{\theta=0}^{2 \pi} e^{-i k \theta} f(\theta).$$
We have $a_0=0$ by the assumption on the integral of $f$. A standard strengthening of the Riemann-Lebesgue lemma says that, if $f$ is $C^r$, then $a_k = o(1/k^r)$. (See, for example, here.) In our case, $f$ is smooth, so we obtain that
$$a_k = O(1/k^r) \ \text{for every}\ r. \tag{1}$$

So
$$\sum_{n=1}^N f(n) = \sum_{k \neq 0} a_k \sum_{n=1}^N e^{ikn} = \sum_{k \neq 0} a_k \frac{e^{ik(N+1)} - e^{ik}}{e^{ik}-1} \tag{2} $$
where there is no $k=0$ term since $a_0=0$.

Now, Zeilberger and Zudlin proved that, for any integers $k$ and $\ell \neq 0$, we have
$$|\pi - k/\ell| \geq c/\ell^{7.11}$$
for some $c>0$. So
$$|k - \ell \pi| \geq c/\ell^{6.11} \geq c'/k^{6.11}$$
for some other constant $c'$. So
$$\frac{e^{ik(N+1)} - e^{ik}}{e^{ik}-1} = O(k^{6.11}). \tag{3}$$

Combining $(1)$ and $(3)$, the summand in (2) is $O(k^{6.11-r})$ for every $r$, and thus in particular is $O(k^{-2})$. So the sum in $(2)$ is bounded, independent of $N$.