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) that the limit is $z_\alpha(x)=0$ for $\alpha<1/2$ and $z_\alpha(x)=x$ for $\alpha>1/2$. I do not have a proof, numerically the $\alpha>1/2$ limit matches very well (see graph below), while the convergence for $\alpha<1/2$ is too slow to make a convincing case.

<IMG SRC="https://ilorentz.org/beenakker/MO/sineiterate_500_0p75.png" WIDTH="400"/>    

Plot of $n^\alpha\sin^{\circ n}(n^{-\alpha}x)$ for $n=500$ and $\alpha=0.75$.