As you've already intuited, this is really a statement about the theory of elliptic curves, not about algebraic geometry proper.
The real statement that you're after, in the sense that I assume you want this as the corollary, is the following: let $(E,p)$ be an elliptic curve then
$$\pi_1^{\acute{e}\text{t}}(E,\overline{p})=T(E)$$
where
$$T(E)=\prod_\ell T_\ell(E)$$
where, here,
$$T_\ell(E)=\varprojlim E[\ell^n]\left(\overline{k}\right)$$
which, in the process, will give you want. Of course, I assume that you're assuming that $K$ is algebraically closed, else we get a canonical isomorphism
$$\pi_1^{\acute{e}t}(E,\overline{p})=T(E)\rtimes G_K$$
where $G_K=\text{Gal}(K^\text{sep}/K)$.
Note that, in particular, this shows that in characteristic $0$ the étale fundamental group is just $\widehat{\mathbb{Z}}^2$ and in positive characteristic it's $(\mathbb{Z}^{(p)})^2\times A$ (here $\mathbb{Z}^{(p)}$ is the product of $\mathbb{Z}_\ell$ for $\ell\ne p$) and $A$ is either $\mathbb{Z}_p$ or $0$ depending whether $E$ is ordinary or supersingular.
(Slight pedantic remark: if one wants to think about Galois representations, one should really replace the $\overline{k}$ above with $k^\text{sep}$. If $\ell\ne p$, where forever $p=\mathrm{char}(k)$, then there is no difference since, as you pointed out, the finite flat group schemes are étale so pick all their points up over the separable closure. But, for us, we need to actually think about the points over the algebraic closure.)
So, assume that $f:C\to E$ is a finite Galois cover of $E$. Then, we immediately deduce many things about $C$: it's a smooth projective connected curve (these all follow from basic considerations about finite étale maps). I claim that $C$ has genus $1$. But, this follows fairly obviously from the Riemann-Hurwitz formula:
$$2g(C)-2=\deg(f)(2(1)-2)+\sum 0=0$$
and thus $g(C)=0$ as desired.
Thus, for a choice $q\in f^{-1}(p)$ we see that $(C,q)$ is an elliptic curve. Moreover, since $f(q)=p$ (by construction) it follows from the basic theory of elliptic curves (a corollary of the ‘Rigidity Lemma’) that $C\to E$ is actually a group map and, in fact, an isogeny.
But, note that $\ker(f)$ is then a subgroup (scheme) of $C$ of degree $\deg(f)$ and since translation by the points of $\ker(f)$ (of which there are actually $\deg(f)$ since $f$, and thus $\ker(f)$ are étale!) are all in $\text{Aut}(C/E)$ we conclude that
$$\text{Gal}(C/E)=\text{Aut}(C/E)^\text{op}=\ker(f)^\text{op}=\ker(f)$$
in particular, $C\to E$ is necessarily abelian.
In particular, take a $p$-Sylow subgroup $P\subseteq \text{Gal}(C/E)$ and consider the associated tower one gets
$$C\to C/P\to E$$
which allows us to, essentially, divide our efforts amongst the $p$-power case, and the prime-to-$p$ case.
The latter is simple. Since
$$f:C\to E$$
is an isogeny, there then exists by the basic theory of elliptic curves, a dual isogeny
$$\widehat{f}:E\to C$$
such that
$$f\circ \widehat{f}=[\deg(f)]$$
as a morphism on $E$. Since we’re in the prime-to-$p$ setting we can conclude that $[\deg(f)]$ is étale and thus we’ve dominated our Galois cover by one of the form $[m]$.
If we’re in the $p$-power case, it’s slightly more annoying but not too bad. Namely, we know that since
$$C\to E$$
is an isogeny, that $E=C/\ker(f)$ and, as we observed above, that $\text{Gal}(C/E)=\ker(f)$. Since $f$ is étale $\ker(f)$ must be a constant subgroup (scheme) of $C$ and so a quotient of $T_p(C)$. But, as before, we can find a dual isogeny
$$\widehat{f}:E\to C$$
and so by the same token, we see that $C=E/M$ for some subgroup (scheme) of $M$, and thus we can identify $T_p(C)$ with $T_p(E)$—the subgroup $\ker(f)\subseteq C$ can really be thought of as a subgroup of $E$. Conversely, we get the full $T_p(E)$ by considering the tower $\{E/E[p^n]^\circ\xrightarrow{f_n}E\}$, where $f_n$ is quotient by $E[p^n]^{\acute{e}\text{t}}$, as $n$-grows. The same sort of uniformization is at play here since we're tacitly identifying
$$(E/E[p^n]^\circ)/(E[p^n]^{\acute{e}\text{t}})=E/E[p^n]=E$$
in our above tower.
EDIT: Most of what I said above is fairly basic algebraic geometry, except, possibly, the dual isogeny statement. So, for self-containment, let me give an easy way to think about such a morphism.
Let's assume for a second that you know that an elliptic curve $(E,e)/k$ is a group scheme in such a way such that for any $L/k$-the group $E(L)$ is functorially identified with $\mathrm{Pic}^0(L)$.
Now, let $\mathscr{C}$ be the category of smooth projective, geometrically connected curves over $k$ with morphisms finite morphisms. Then, it's well-known (by the valuative criterion) that for any object $C\in\mathscr{C}$ we have that
$$\text{Hom}(C,E)=E(K(C))=\mathrm{Pic}^0(E_{K(C)})$$
in particular, we see that to define a group map $E\to E'$ of elliptic curves, it suffices to define a group map
$$\mathrm{Pic}^0(E(K(C))=E(C)\to E'(C)=\mathrm{Pic}^0(E'(K(C))$$
functorially in $C$.
So, suppose that we're given a morphism $f:E\to E'$ in $\mathscr{C}$ which is, in fact, a group map. We then want to define a map $\widehat{f}:E'\to E$ such that $f\circ \widehat{f}=[\deg(f)]$. To do this, we use the above observation. Namely, for all $C\in\mathscr{C}$ let us define
$$\widehat{f}:\mathrm{Pic}(E'_{K(C))})\to \mathrm{Pic}(E_{K(C)})$$
by
$$\mathscr{L}\mapsto f^\ast(\mathscr{L})$$
it's evident that this is a functorial group map and thus we obtain a group morphism
$$\widehat{f}:E'\to E$$
as desired.
To see that it satisfies the desired property that $f\circ \widehat{f}=[\deg(f)]$ it suffices to see that they agree on $\overline{k}$-points. But, note that if $\mathscr{L}=\mathcal{O}(p-e)\in\mathrm{Pic}^0(E'_{\overline{k}})$, then
$$f^\ast\mathscr{L}=\mathcal{O}(D)$$
where
$$D=(p-e)\times_{E'}E=\sum_{q\in f^{-1}(p)}e_q q+\sum_{q'\in f^{-1}(e))}e_{q'}q'$$
and then so, applying $f$ to this we get
$$\begin{aligned}f\left(\sum_{q\in f^{-1}(p)}e_q q+\sum_{q'\in f^{-1}(e)}e_{q'}e\right) &=\sum_{q\in f^{-1}(q)}e_q f(q)+\sum_{q'\in f^{-1}(e)}e_{q'}f(q')\\ &=\sum_{q\in f^{-1}(p)}e_q p+\sum_{q'\in f^{-1}(e)}e_{q'}e\\ &= \deg(f) p-\deg(f) e\end{aligned}$$
which proves the claim.
