In dimensions up to three, every manifold is triangulable (this is classical). In dimension 4, there are simply connected non-triangulable manifolds (such as the E8 manifold); in fact, a closed 4-manifold is triangulable if and only if it's smoothable. (Pick a triangulation; the links of vertices are always both homology and homotopy spheres, thus by the Poincare conjecture are $S^3$, so this gives a PL structure, and thus is smoothable. This is probably true more generally than closed manifolds, but I'm feeling a little paranoid.)

The crucial papers pre-Manolescu are, firstly, [Galewski-Stern](http://www.maths.ed.ac.uk/~aar/papers/galester.pdf) and Matumoto's thesis (which I don't have a link to): these prove that every closed manifold if dimension $n \geq 5$ are triangulable if and only if there is a homology 3-sphere $Y$ such that $Y \# Y$ bounds a homology ball, and the Rokhlin invariant $\mu(Y) = 1 \in \Bbb Z/2\Bbb Z$. This is what Manolescu disproved (there is no such 3-manifold); the Galewski-Stern paper clarifies what dimensions their theorem is proved for after you start dropping assumptions about compactness and the boundary.

But importantly for your question is Galewski and Stern's [sequel paper](http://faculty.sites.uci.edu/rstern/files/2011/03/13_Universal_Tri.pdf). Their theorem 2.1 (plus Manolescu's result) implies that there are non-triangulable manifolds in every dimension $n \geq 5$. However, all *orientable* 5-dimensional manifolds are triangulable. In dimensions at least 6, though, you can use their construction to produce non-triangulable orientable manifolds.