Specific question: is there a finite CW complex homeomorphic to a sphere such that one of its maximal cells has as its closure a ball whose boundary is embedded in the CW-sphere as an Alexander horned sphere?
1) No. If $K$ is the $2$-skeleton of the CW complex, then $K$ has a mapping cylinder neighborhood in $S^3$, and hence by Nicholson's theorem it is tame, that is, equivalent to a subpolyhedron of $S^3$ by a homeomorphism $h$ of $S^3$. Since $K$ is $2$-dimensional, it is not hard to show that $h$ also takes any $2$-sphere in $K$ onto a subpolyhedron of $S^3$. So $K$ cannot contain the horned sphere.
2) Yes, if you allow an unspecified wild codimension one sphere in place of the Alexander horned sphere. By Example 7.11.2 on p.419 in the Daverman-Venema book (this seems to be among a few original results in the book, perhaps the most important one), $S^n$ for $n\ge 6$ contains a wildly embedded sphere $\Sigma$ with a mapping cylinder neighborhood $N$. By construction, the complement to the interior of $N$ consists of two closed $n$-balls. It follows that the closures of the complementary domains of $\Sigma$ are the mapping cones of some self-maps of $S^{n-1}$. So we get a CW-complex with one $0$-cell, one $(n-1)$-cell and two $n$-cells, which is homeomorphic to $S^n$, and has a wild $(n-1)$-skeleton.
Follow-up question: if the answer is no, is there a finite CW complex homeomorphic to a sphere such that the closure of one of the maximal cells is a ball, but the closure of its complement is not a ball?
1) Yes. You can glue one of the complementary domains of $\Sigma$ and an $n$-ball along their boundary sphere. The result is again $S^n$, according to Proposition 7.10.1 in Daverman-Venema.
2) I should also mention a simpler but somewhat similar example. Using the Edwards-Cannon theorem, it is not hard to construct a finite regular CW-complex $K$ that is homeomorphic to $S^5$, even though the boundary of some $2$-cell (in fact, of each $2$-cell) of $K$ is wild, viewed as a copy of $S^1$ in $S^5$. In more detail, if $H$ is a traingulation of a non-simply-connected homology $3$-sphere, then the double suspension $S^0*S^0*H$ is a simplicial complex homeomorphic to $S^5$; the desired CW-complex is the 'prejoin' $(S^0*S^0)+H$, which is PL homeomorphic to $S^0*S^0*H$ and has all its $2$-cells attached to the suspension circle $S^0*S^0$. On identifying regular CW-complexes with their posets of nonempty faces, the prejoin $P+Q$ of two posets is defined by placing all the elements of $P$ below all the elements of $Q$ in the Hasse diagram, and keeping the original order within $P$ and within $Q$. The order complex of $P+Q$ is easily seen to be isomorphic to the join of the order complexes of $P$ and of $Q$. As an example, $S^0*S^0*pt$ is a simplicial complex with $4$ of $2$-simplices, whereas $(S^0*S^0)+pt$ is a cell complex with only one (quadrilateral) $2$-cell.
Beware that $K$ itself is a PL CW-complex, with PL attaching maps; the only trouble is with the homeomorphism between $K$ and $S^5$.
