This sparsity holds even without the assumptions that $p>2$ and $2(p-1)>n^2$. (Those are used to get a horizontal vanishing line). As in the comments, it comes down to the fact that you can use $\widehat{E(n)}\simeq L_{K(n)}E(n)$ in place of $E_n$, and $\widehat{E(n)}$ has homotopy groups concentrated in degrees that are a multiple of $2(p-1)$.
One way to get sparsity directly from the group cohomology is to use that there's a copy of $\mathbb{Z}_p^\times$ sitting inside the center of $\mathbb{G}_n$, and further sitting inside this is the subgroup $\mathbb{F}_p^\times\subset\mathbb{Z}_p^\times$ of roots of unity (the Teichmüller lifts). Because the order of $\mathbb{F}_p^\times$ is invertible in $\pi_\ast E_n$, you get an isomorphism $$ H^\ast_c(\mathbb{G}_n;\pi_\ast E_n)\cong H^\ast_c(\mathbb{G}_n/\mathbb{F}_p^\times;(\pi_\ast E_n)^{\mathbb{F}_p^\times}). $$ An element $\ell\in\mathbb{Z}_p^\times$ corresponds to multiplication by $\ell$ on the formal group of $E_n$, so acts on $\pi_{2t}E_n$ as multiplication by $\ell^t$. This implies $$ (\pi_t E_n)^{\mathbb{F}_p^\times} = \begin{cases} \pi_t E_n &2(p-1)\mid n\\0&\text{otherwise}, \end{cases} $$ and thus $H^\ast_c(\mathbb{G}_n;\pi_t E_n) = 0$ unless $2(p-1)\mid n$.