Is the E8 manifold homeomorphic to a CW complex?

Is the E8 manifold homeomorphic to a CW complex? (I know that it is not triangulable)

Edit: The E8 manifold is the unique compact (without boundary), simply connected topological 4-manifold, whose 2nd integer homology (equipped with its intersection form) is isomorphic to the $E_8$ lattice. A detailed description of this manifold can be found in an answer to this MSE question.

It appears to be an open problem to determine if every $n$-dimensional topological manifold admits structure of a CW complex. Positive answer (due to hard work of many people, most importantly - Kirby and Siebenmann for $n\ge 6$, Freedman and Quinn for $n=5$) is known in all dimensions except for $n=4$. Furthermore, all connected noncompact 4-dimensional manifolds are known to be triangulable (this is due to Quinn) and, hence, the answer is positive in this case as well.

Given that the E8-manifold is, in some sense, the simplest nonsmoothable 4-dimensional manifold, it makes sense to ask if it is homeomorphic to a CW complex.

I think, we can safely treat Andrew Ranicky as an expert in these matters. In a comment to his answer here Ranicky asked the following:

Question. Is it true that a 4-dimensional manifold $X$ admits a CW-complex structure if and only if $X$ is smoothable?

Since the E8-manifold is not smoothable, your question is a special case of an open (and, I guess, important) problem in 4-dimensional topology.