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Your guess is correct, and I think this is a confusion pretty much everyone has when they first see these computations. What's going on here is essentially just the simple algebraic fact that $$\mathbb{Z}[p^{-1}][[x]]\neq\mathbb{Z}[[x]][p^{-1}].$$

In computation (1), you can work directly with a Gysin sequence for $E_n$, without going through $K(n)$ and Bocksteins. The Gysin sequence tells you that $$E_n^*(B\mathbb{Z}/p)=E_n^*[[x]]/([p](x)),$$ where $[p](x)$ is the $p$-series of the formal group law. More generally, this holds for any complex-oriented theory $E$ such that $[p](x)\in E^*[[x]]$ is a nonzero divisor.

Since the formal group law for $E_n$ has height $n$, the first coefficient of $[p](x)$ which is a unit is the $x^{p^n}$ coefficient. It follows that $[p](x)$differs from a monic polynomial of degree $p^n$ by a unit in $E_n^*[[x]]$, so $E_n^*(B\mathbb{Z}/p)$ is free of rank $p^n$ over $E_n^*$.

On the other hand, if we do the same argument computation with $E_n\otimes \mathbb{Q}$ instead of $E_n$, $$\frac{[p](x)}{x}=p+\dots\in(E_n\otimes \mathbb{Q})^*[[x]]$$ is now a unit. Thus $(E_n\otimes\mathbb{Q})^*(B\mathbb{Z}/p)$ is free of rank 1 over $(E_n\otimes\mathbb{Q})^*$. However, $[p](x)/x$ is only a unit in $(E_n\otimes \mathbb{Q})^*[[x]]$, not in $E_n^*[[x]]\otimes\mathbb{Q}$. This is because to construct a power series inverse for it, we need coefficients with arbitrarily large powers of $p$ in the denominator. Thus we find that $(E_n\otimes \mathbb{Q})^*[[x]]/([p](x))$ and $E_n^*[[x]]\otimes\mathbb{Q}/([p](x))$ look quite different from each other, and this is exactly the discrepancy between your two computations.

As for a computation along the lines of (2) that gives the answer from (1), I don't know of anything like that. You could make an AHSS computation for $E_n$ rather than $E_n\otimes\mathbb{Q}$ and only tensor with $\mathbb{Q}$ afterwards, which would give the right answer. However, the AHSS $$H^*(\mathbb{Z}/p,E_n^*)\implies E_n^*(B\mathbb{Z}/p)$$ is quite subtle: I don't know how to compute the differentials in it without already knowing the final answer, and the filtration of the spectral sequence distorts almost all of the structure of $E_n^*(B\mathbb{Z}/p)$ (for instance, the associated graded you get from the spectral sequence consists mostly of $p$-torsion, whereas the actual answer is torsion-free).

1

Your guess is correct. What's going on here is essentially just the simple algebraic fact that $$\mathbb{Z}[p^{-1}][[x]]\neq\mathbb{Z}[[x]][p^{-1}].$$

In computation (1), you can work directly with a Gysin sequence for $E_n$, without going through $K(n)$ and Bocksteins. The Gysin sequence tells you that $$E_n^*(B\mathbb{Z}/p)=E_n^*[[x]]/([p](x)),$$ where $[p](x)$ is the $p$-series of the formal group law. More generally, this holds for any complex-oriented theory $E$ such that $[p](x)\in E^*[[x]]$ is a nonzero divisor.

Since the formal group law for $E_n$ has height $n$, the first coefficient of $[p](x)$ which is a unit is the $x^{p^n}$ coefficient. It follows that $[p](x)$differs from a monic polynomial of degree $p^n$ by a unit in $E_n^*[[x]]$, so $E_n^*(B\mathbb{Z}/p)$ is free of rank $p^n$ over $E_n^*$.

On the other hand, if we do the same argument with $E_n\otimes \mathbb{Q}$ instead of $E_n$, $$\frac{[p](x)}{x}=p+\dots\in(E_n\otimes \mathbb{Q})^*[[x]]$$ is now a unit. Thus $(E_n\otimes\mathbb{Q})^*(B\mathbb{Z}/p)$ is free of rank 1 over $(E_n\otimes\mathbb{Q})^*$. However, $[p](x)/x$ is only a unit in $(E_n\otimes \mathbb{Q})^*[[x]]$, not in $E_n^*[[x]]\otimes\mathbb{Q}$. This is because to construct a power series inverse for it, we need coefficients with arbitrarily large powers of $p$ in the denominator. Thus we find that $(E_n\otimes \mathbb{Q})^*[[x]]/([p](x))$ and $E_n^*[[x]]\otimes\mathbb{Q}/([p](x))$ look quite different from each other, and this is exactly the discrepancy between your two computations.

As for a computation along the lines of (2) that gives the answer from (1), I don't know of anything like that. You could make an AHSS computation for $E_n$ rather than $E_n\otimes\mathbb{Q}$ and only tensor with $\mathbb{Q}$ afterwards, which would give the right answer. However, the AHSS $$H^*(\mathbb{Z}/p,E_n^*)\implies E_n^*(B\mathbb{Z}/p)$$ is quite subtle: I don't know how to compute the differentials in it without already knowing the final answer, and the filtration of the spectral sequence distorts almost all of the structure of $E_n^*(B\mathbb{Z}/p)$ (for instance, the associated graded you get from the spectral sequence consists mostly of $p$-torsion, whereas the actual answer is torsion-free).