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A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series, 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.

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.

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

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 $(\text{sign}\,x)\,\sqrt{3}\,n^{\alpha-1/2}$.

A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series by C. Towse (2014), 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.

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.

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

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 $(\text{sign}\,x)\,\sqrt{3}\,n^{\alpha-1/2}$.

A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series, 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.

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.

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 $\alpha>1/2$. 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.

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

$^\ast$ For $\alpha<1/2$ we can proceed as follows: Assume a power law decay, $y_n=n^\alpha\sin^{\circ n}(n^{-\alpha}x)=cn^{-p}$, substitute into $y_{n+1}=n^\alpha \sin(y_n/n^\alpha)$, and expand for large $n$. So we have $$c(n+1)^{-p}\approx cn^{-p}-cpn^{-p-1},\;\; n^\alpha\sin(cn^{-p-\alpha})\approx cn^{-p}-\tfrac{1}{6}c^3n^{-3p-2\alpha},$$ and equating these two expressions gives $p=\tfrac{1}{2}-\alpha$, $c=\sqrt{6p}=\sqrt{3-6\alpha}$. We thus arrive at the large-$n$ asymptotics $$n^\alpha\sin^{\circ n}(n^{-\alpha}x)\rightarrow \sqrt{3-6\alpha}\,n^{\alpha-1/2}\rightarrow 0,\;\;\text{for}\;\;\alpha<1/2.$$ Numerically, this agrees quite well for $\alpha\lesssim 0.1$, see plot, for larger $\alpha$ the curve remains flat but at a larger value, presumably because of higher order terms.

Plot of $n^\alpha\sin^{\circ n}(n^{-\alpha}x)$ for $n=500$ and $\alpha=0.1$. The horizontal straight line is the asymptote $\sqrt{3-6\alpha}\,n^{\alpha-1/2}$.

A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series, 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.

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.

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

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 $(\text{sign}\,x)\,\sqrt{3}\,n^{\alpha-1/2}$.

A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series, 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.

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.

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.

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

A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series, 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.

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.

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 $\alpha>1/2$. 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.

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

$^\ast$ For $\alpha<1/2$ we can proceed as follows: Assume a power law decay, $y_n=n^\alpha\sin^{\circ n}(n^{-\alpha}x)=cn^{-p}$, substitute into $y_{n+1}=n^\alpha \sin(y_n/n^\alpha)$, and expand for large $n$. So we have $$c(n+1)^{-p}\approx cn^{-p}-cpn^{-p-1},\;\; n^\alpha\sin(cn^{-p-\alpha})\approx cn^{-p}-\tfrac{1}{6}c^3n^{-3p-2\alpha},$$ and equating these two expressions gives $p=\tfrac{1}{2}-\alpha$, $c=\sqrt{6p}=\sqrt{3-6\alpha}$. We thus arrive at the large-$n$ asymptotics $$n^\alpha\sin^{\circ n}(n^{-\alpha}x)\rightarrow \sqrt{3-6\alpha}\,n^{\alpha-1/2}\rightarrow 0,\;\;\text{for}\;\;\alpha<1/2.$$ Numerically, this agrees quite well for $\alpha\lesssim 0.1$, see plot, for larger $\alpha$ the curve remains flat but at a larger value, presumably because of higher order terms.

Plot of $n^\alpha\sin^{\circ n}(n^{-\alpha}x)$ for $n=500$ and $\alpha=0.1$. The horizontal straight line is the asymptote $\sqrt{3-6\alpha}\,n^{\alpha-1/2}$.