I think in every dimension $\ge 4$ there is a fake real projective space, i.e., a manifold that is homotopy equivalent but not homeomorphic to $RP^n$.
Here you can find a computation for the topological surgery structure set of $RP^n$. It is stated for $n>4$ but I think it extends to $n=4$ because $\mathbb Z_2$ is good in the sense of Freedman. To get a homeomorphism classification of homotopy $RP^n$'s we also need to know the group of homotopy self equivalences of (any manifold homotopy equivalent to) $RP^n$: it is $\mathbb Z_2$ if $n$ is odd and trivial if $n$ is even (see e.g. p.61 of Rutter's survey ``Spaces of homotopy self-equivalences'').
Let $f_i: M_i\to RP^n$ be homotopy equivalences representing different elements in the structure set. If $d_{ij}: M_i\to M_j$ is a homeomorphism, then $f_i^{-1}f_j d_{ij}$ is a homotopy self-equivalence of $M_i$. Thus if $n$ is even, then $f_i^{-1}f_j d_{ij}$ is homotopic to the identity, and hence $f_i$, $f_j$ are the same in the structure set, i.e., $i=j$. Similarly, if $n$ is odd, $f_i^{-1}f_j d_{ij}$ and $f_i^{-1}f_k d_{ik}$ are homotopic to a nontrivial self-equivalence, then $f_j$, $f_k$ are the same in the structure set. The computation linked above says that the structure set has $2$ elements only if $n=4$ which is even. Thus in every dimension there is a fake $RP^n$.
A word of caution: the above argument does not show that the fake $RP^n$ is smoothable. For example, the only fake $RP^4$ is not smmothable as I learned from ``Invariant knots of free involutions of $S^4$'' by Ruberman, see here. I do believe that there exist many smoothable fake $RP^n$'s.