I was under the impression that any hyperbolic manifold (finite volume or not) had $\mathbb{H}^n$ as its universal cover, but on wikipedia it is suggested that this only true for closed hyperbolic manifolds.
In that case, what is the universal cover of a noncompact hyperbolic manifold, such as an infinite volume hyperbolic manifold?
The important assumption is completeness, not compactness. The key fact is that every complete, connected, simply connected hyperbolic manifold is isometric to $\mathbb{H}^n$.
This implies that the universal cover of any complete, connected hyperbolic manifold is isometric to $\mathbb{H}^n$, since a Riemannian cover of a complete manifold must also be complete (see this question). In particular, this is also true for compact connected hyperbolic manifolds, since compact Riemannian manifolds are complete.
Edit: Thanks to Lee Mosher for pointing out in the comments that it is not true that the universal cover of any connected hyperbolic manifold is isometric to an open submanifold of $\mathbb{H}^n$.