A rescaling is needed for a nontrivial limit. As discussed in <A HREF="https://www.jstor.org/stable/10.4169/math.mag.87.5.338#metadata_info_tab_contents">Iteration of Sine and Related Power Series</A>, denoting the $n$-th iterate by $\sin^{\circ n}x$, one has the limit
$$\lim_{n\rightarrow\infty}\sqrt n\sin^{\circ n}(x/\sqrt n)=\frac{x}{\sqrt{1+x^2/3}}.$$
The graph (from the cited paper) shows that the limit is attained quite rapidly.
<IMG SRC="https://ilorentz.org/beenakker/MO/sineiterate_1.png"/>

Without the rescaling the iterated sine converges to zero (for the reasons indicated in the comments to the OP). The convergence is slow, see the graph.

<IMG SRC="https://ilorentz.org/beenakker/MO/sineiterate_2.png"/>

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For the general rescaling,
$$z_\alpha(x)=\lim_{n\rightarrow\infty}n^\alpha\sin^{\circ n}(n^{-\alpha}x),$$
I surmise (based on the small-$x$ expansion of the sine$^\ast$) that the limit is $z_\alpha(x)=0$ for $\alpha<1/2$ and $z_\alpha(x)=x$ for $\alpha>1/2$. This agrees quite well with the numerics, see plots below

<IMG SRC="https://ilorentz.org/beenakker/MO/sineiterate_500_0p75.png" WIDTH="300"/><IMG SRC="https://ilorentz.org/beenakker/MO/sine_iterate_500_0p25.png" WIDTH="300"/>    

Plots of $n^\alpha\sin^{\circ n}(n^{-\alpha}x)$ for $n=500$ and $\alpha=0.75$ (left) and $\alpha=0.25$ (right). The horizontal lines in the right plot show the asymptote $(\text{sign}\,x)\,\sqrt{3}\,n^{\alpha-1/2}$.

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$^\ast$ For $\alpha<1/2$ we can proceed as follows: Assume a power law decay, $y_n=(\text{sign}\,x)\,n^\alpha\sin^{\circ n}(n^{-\alpha}x)=(\text{sign}\,x)cn^{-p}$, substitute into $y_{n+1}=(n+1)^\alpha \sin(y_n/n^\alpha)$, and expand for large $n$. So we have
$$c(n+1)^{-p}\approx cn^{-p}-cpn^{-p-1},\;\; (n+1)^\alpha\sin(cn^{-p-\alpha})\approx cn^{-p}+\alpha cn^{-p-1}-\tfrac{1}{6}c^3n^{-3p-2\alpha},$$
and equating these two expressions gives $p=\tfrac{1}{2}-\alpha$, $c=\sqrt{6(p+\alpha)}=\sqrt{3}$. We thus arrive at the large-$n$ asymptotics
$$n^\alpha\sin^{\circ n}(n^{-\alpha}x)\rightarrow (\text{sign}\,x)\,\sqrt{3}\,n^{\alpha-1/2}\rightarrow 0,\;\;\text{for}\;\;\alpha<1/2.$$
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