I'm confused about a seeming contradiction that is probably just a reflection of ignorance on my part. Let's try to compute the Morava E-theory of $B \mathbb{Z}/p$ in two different ways.

First, following Ravenel-Wilson (see also Hopkins-Kuhn-Ravenel, Hunton, etc.), one computes (via knowledge of $K(n)^\ast(\mathbb{C} P^\infty)$ and a Gysin sequence) that the Morava K-theory of $B \mathbb{Z}/p$ is $$K(n)^\ast (B \mathbb{Z}/p) = K(n)_* [x] / x^{p^n}$$ where $x$ is of degree $2$. In particular, it is concentrated in even degrees and free of rank $p^n$ over $K(n)_\ast$. By Bockstein arguments, the Morava E-theory is then also free of the same rank.

Second, using the knowledge that $E_n^\ast (B \mathbb{Z}/p)$ is free over $E_n^\ast$, it must embed into its rationalisation, $E_n^\ast (B \mathbb{Z}/p) \otimes \mathbb{Q}$. Let's try to compute the rank of that rationalisation using the group cohomological Atiyah-Hirzebruch spectral sequence: $$H^*(\mathbb{Z} / p, E_n^\ast \otimes \mathbb{Q}) \implies (E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$$ However, the $E_2$ term of the spectral sequence is rank 1 over $E_n^\ast \otimes \mathbb{Q}$ since $\mathbb{Z} / p$ is a finite group, and must collapse there. Therefore it certainly doesn't contain a sublattice of rank $p^n$.

I've indicated my misgivings with the second argument by notating the target of the spectral sequence as $(E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$, that is, the value on $B \mathbb{Z}/p$ of the cohomology theory which is the rationalisation of $E_n$ (whose homotopy groups are $E_n^\ast \otimes \mathbb{Q}$). My guess is that this is NOT the same as the rationalisation of the E-theory of $B \mathbb{Z}/p$.

So my real question is: is it easy to see why this is the case? More importantly, can one compute the correct answer (i.e., coming from argument 1) through methods based upon argument 2? I'm envisioning some sort of spectral sequence that knows how rationalisation and cohomology may fail to commute.