2 Multiplication by p should be purely inseparable, not just inseparable...

In characteristic $p$, every map $E_1 \to E_2$ factors as a power of the Frobenius $\varphi_r \colon E_1 \to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} \to E_2$, and we find $r$ by looking at the inseparable degree of our map (if the map is separable, then $r=0$, as Pete pointed out).

Now, in the case of interest, if $E$ is supersingular, $\widehat{\varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being purely inseparable). But then $\widehat{\varphi} \colon E^{(p)} \to E$ factors as $E^{(p)} \to E^{(p^2)} \to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism.

It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$.

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In characteristic $p$, every map $E_1 \to E_2$ factors as a power of the Frobenius $\varphi_r \colon E_1 \to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} \to E_2$.

Now, in the case of interest, if $E$ is supersingular, $\widehat{\varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being inseparable). But then $\widehat{\varphi} \colon E^{(p)} \to E$ factors as $E^{(p)} \to E^{(p^2)} \to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism.

It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$.