Using $\sin nz=\frac{1}{2i}(e^{inz}-e^{-inz})$ works. The constant is irrelevant for the convergence.
Deal with the two exponentials separately. Let $z=x+iy$. Then $|e^{inz}|=e^{-ny}$, and $|e^{i((n+1)z}|=e^{-(n+1)y}$.
Thus, remembering about the $2^n$ in the denominator, we see that the norm of the ratio of two consecutive terms is $\frac{e^{-y}}{2}$. This norm is $\lt 1$ precisely if $e^{-y} \lt 2$, that is, if $y\gt -\log 2$.
In the same way, for the term in $e^{-iz}$, the norm of the ratio of two consecutive terms is $\lt 1$ precisely if $y \lt \log 2$. Thus, by the Weierstrass $M$-test, we have analyticity if $-\log 2\lt y\lt \log 2$.