Note that if $\pi$ were rational (with even numerator), then $\sin(n)$ would equal $1$ periodically, so the series would diverge.  Similarly if $\pi$ were a sufficiently strong [Liouville number][1].  Thus, to establish convergence, one must use some quantitative measure of the irrationality of $\pi$.

It is known that the [irrationality measure][2] $\mu$ of $\pi$ is finite (indeed, the current best bound is $\mu \leq 7.60630853$).  Thus, one has a lower bound
$$ | \pi - \frac{p}{q} | \gg \frac{1}{q^\mu}$$
for all $p,q$.  This implies that
$$ \| p/\pi \|_{{\bf R}/{\bf Z}} \gg \frac{1}{p^{\mu-1}},$$
for all $p$, where $\|x\|$ denotes the distance to the nearest integer.  In particular, if $I \subset {\bf R}/{\bf Z}$ is an arc of length $0 < \delta < 1$, the set of $n$ for which $n/\pi \hbox{ mod } 1 \in I$ is $\gg \delta^{-1/(\mu-1)}$-separated.  This implies, for any natural number $k$, that the number of $n$ in $[2^k,2^{k+1}]$ such that $|\sin(n)|$ lies in any given interval $J$ of length $2^{-k}$ (which forces $n/\pi \hbox{ mod } 1$ to lie in the union of at most four intervals of length at most $O(2^{-k/2})$) is at most $\ll 2^{k(1 - \frac{1}{2(\mu-1)})}$.  Noting (from Taylor expansion) that $|\sin(n)|^n \ll \exp( - j)$ if $n \in [2^k,2^{k+1}]$ and $|\sin(n)| \in [1 - \frac{j+1}{2^k}, 1-\frac{j}{2^k}]$, we conclude on summing in $j$ that
$$ \sum_{2^k \leq n < 2^{k+1}} |\sin(n)|^n \ll 2^{k(1 - \frac{1}{2(\mu-1)})}$$
and hence 
$$ \sum_{2^k \leq n < 2^{k+1}} \frac{|\sin(n)|^n}{n} \ll 2^{- k\frac{1}{2(\mu-1)}}.$$
The geometric series on the RHS is summable in $k$, so the series $\sum_{n=1}^\infty \frac{|\sin(n)|^n}{n}$ is convergent.


  [1]: https://en.wikipedia.org/wiki/Liouville_number
  [2]: https://en.wikipedia.org/wiki/Liouville_number#Irrationality_measure