As I understand it, the BTZ black hole metric has a constant curvature, and hence the space is locally isometric to AdS$_3$. Moreover, the global space can be written as a quotient space of AdS$_3$ by a discrete group. However, I don't see a simple way of seeing this from the coordinated in the question.

The BTZ black hole metric has a constant curvature, and hence the space is locally isometric to AdS$_3$. Moreover, the global space can be written as a quotient space of AdS$_3$ by a discrete group. However, I don't see a simple way of seeing this from the coordinated in the question.

That the space is asymptotically AdS$_3$ means that it satisfies the Brown-Henneaux boundary conditions. Change coordinates to $\rho = e^r$. Then the asymptotic metric in the question takes the form $$ ds^2 = l^2 d\rho^2 + e^{2\rho}\left(-\frac{1}{l^2} dt^2 + d\phi^2\right) - J\,dt\,d\phi $$ The general Fefferman- Graham expansion takes the form $$ ds^2 = l^2 d\rho^2 + \left(e^{2\rho} \gamma^{(0)}_{ij} + \rho \gamma^{(1)}_{ij} + \gamma^{(2)}_{ij} + \dotsb \right) dx^i dx^j $$ but for a metric satisfying the Brown-Henneaux boundary conditions the $\gamma^{(1)}$ term vanishes. This is enough to ensure that the asymptotic isometry algebra is the same as in AdS$_3$, namely that of two Virasoro algebras.