I think the statement of the question is pretty straightforward. Given a finite $n$-dimensional CW complex, are there necessary and sufficient conditions for determining that it is also a compact $n$-dimensional topological manifold (possibly with boundary)? Most of the questions I have found are about the converse, "When is a topological manifold a CW-complex?" so I thought it would be useful to consider the other side of the picture.

First of all, we already have 2nd countable and Hausdorff so we only need to determine when it is locally Euclidean. One necessary condition is that every point must be contained in the closure of at least one $n$-cell. One the other hand, a point can't be in too many $n$-cells, as the wedge of spheres is not a manifold. But these criteria are clearly not sufficient (or very precise: what does "too many" mean exactly?).

Another necessary condition that I think of right away is that the space must satisfy Poincare duality (with $\mathbb{Z}/2$ coefficients). I'm not sure if this is practical at all, but maybe it is useful if you are working with an explicit cell-structure and a concrete description of the cellular chain complex.

After that, I don't know how to proceed. I am assuming this is a difficult problem, since Google searches haven't answered my question yet. One idea that I have is to try and come up with an "obstruction" to this, where cell-by-cell we determine if every point in that cell admits an $n$-dimensional Euclidean nhd, and we have that the CW-complex is a manifold iff this "obstruction," computed at the homology/cohomology level, vanishes.

Any suggestions (or references to a solution) will be appreciated.