Suppose a finite group G acts freely and continuously on an ndimensional CWcomplex X. Then can we conclude that the orbit space of this action is still an ndimensional CWcomplex? (or homotopy equivalent to an ndimensional CWcomplex?) In particular, we do not assume G acts cellularly on X.

Lemma If $X$ is a countable locally finite CWcomplex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CWcomplex. Proof Any metrizable ANR is homotopy equivalent to a CWcomplex (I am not sure who proved it first but see Theorem 3.6.1 here. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$ projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired. Remark In seeing whether $X/G$ is homeomorphic to a CWcomplex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CWcomplexes (KirbySiebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$). So there might exist manifolds not homeomorphic to CWcomplexes but whose finite covers are PL. 


The 3sphere gives an example of an action with fixed points. If one takes the solid Alexander horned sphere, then Bing proved that its double is homeomorphic to the 3sphere. So the quotient of the involution acting on $S^3$ is the solid Alexander horned sphere. However, the solid horned sphere is not homeomorphic to a CW complex. This follows from the answer to this question on the Alexander horned sphere. If the solid Alexander horned sphere were a CW complex, then one could attach the exterior 3ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3cell, which is a contradiction to the other question. 


If $G$ (finite or more generally discrete) acts cellularly on $X$, i.e.
then $X/G$ is a CWcomplex. This follows from Prop. 1.15 and Ex. 1.17(2) of tom Dieck: Transformation Groups 


This is not really an answer, but a comment about an interesting special case. Suppose that $G$ acts smoothly on $S^2$. By averaging we can choose a $G$invariant Riemannian metric. This gives $S^2$ a conformal structure, making it a Riemann surface. Any Riemann surface homeomorphic to $S^2$ is conformally equivalent to the standard Riemann sphere. Thus, we can reduce to the case where $G$ acts on $\mathbb{C}\cup\{\infty\}$ by conformal and anticonformal maps, which must have the form $z\mapsto (az+b)/(cz+d)$ or $z\mapsto (a\overline{z}+b)/(c\overline{z}+d)$. I think it even works out here that the quotient $(\mathbb{C}\cup\{\infty\})/G$ is always either a sphere or a disc. Thus, one cannot get any local pathology in this context. This contrasts with other settings where smooth functions can generate topological pathology: for example, any closed subset of $\mathbb{R}^n$, however fractal, can be expressed as the zero set of a smooth function $f\colon\mathbb{R}^n\to\mathbb{R}$. Along somewhat similar lines, I think one can show that when $X$ is a onedimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a onedimensional CW complex (up to homeomorphism, not just homotopy equivalence). 

