This limit converges to $\frac{\sqrt3}2$. The idea is that $\sin(x) = x - \frac{x^3}6 + O(x^5)$, so we start with $\frac1{\sqrt n}$ and repeatedly subtract $\frac{x^3}6$. We can approximate this discrete system with the differential equation $f'(x) = -\frac{f(x)^3}6$, which has a solution $f(x)=\frac{\sqrt3}{\sqrt{c + x}}$, and we want $f(0) = \frac1{\sqrt n}$, so $c = 3n$, and then $f(n) = \frac{\sqrt3}{2\sqrt n}$.

For a formal proof, we can define two sequences $a_0 = \frac1{\sqrt n}, a_{i+1} = \sin(a_i)$ and $b_i = \frac{\sqrt3}{\sqrt{3n + i}}$. We have $b_{i+1} = b_i \frac{\sqrt{3}}{\sqrt{3 + b_i^2}}$. Let's bound $c_i = |a_i - b_i|$. By looking at the Taylor series, we have $|\sin(b_{i-1}) - b_i| = O(b_i^5)$. Additionally, $|a_i - \sin(b_{i-1})| = |\sin(a_{i-1}) - \sin(b_{i-1})| \leq |a_{i-1} - b_{i-1}|$, so $c_i \leq c_{i-1} + O(b_i^5) \leq c_{i-1} + O(n^{-\frac52})$, or $c_i \leq O(n^{-\frac32})$, so the approximation given by $b_n$ for $a_n$ is good enough.