The series diverges. The sloppy reasoning is that $|\sin(n^2)|$ is roughly a random value between $0$ and $1$ with an average greater than or equal to $0.1$ and $\sum_n \frac{0.1}{n}$ certainly diverges.
To make this more rigorous, as @H.H.Rugh points out in his solution, as $n \to \infty$, $n^2$ is equidistributed mod $\pi$ (using Corollary 6 from here). Thus
$$|\sin(n^2)| > \frac{1}{2} \ \text{ whenever }\ n^2 \text{ mod }\pi \in \left(\frac{\pi}{6},\frac{5\pi}{6}\right)$$
which happens two thirds of the time (asymptotically). Therefore for large enough $N \in \mathbb{Z}$,
$$\sum_{n=N+1}^{2N} \frac{|\sin(n^2)|}{n}> \frac{1}{2N} \sum_{n=N+1}^{2N} |\sin(n^2)| > \frac{1}{2N} \frac{N}{2} \frac{1}{2} = \frac{1}{8}$$
since $|\sin(n^2)| > \frac{1}{2}$ for more than one half of the $n$ in $\{N+1,...,2N\}$
Therefore the series diverges since the tail of the series diverges:
$$\sum_{n=N+1}^{\infty} \frac{|\sin(n^2)|}{n} = \sum_{k=0}^\infty \left(\sum_{n=2^k N+1}^{2^{k+1}N} \frac{|\sin(n^2)|}{n} \right)> \sum_{k=0}^\infty \frac{1}{8} = \infty$$