What is needed to make Euclidean spaces isomorphic as groups?

Consider the abelian groups $G_n=(\mathbb R^n,+)$ for $n\geq1$.

Claim: For any $n$ and $m$ the groups $G_n$ and $G_m$ are isomorphic.

This claim is true if one assumes the axiom of choice, and I have sketched a proof below. But this claim seems much weaker than the axiom of choice, which brings me to the following questions:

1. Are there any milder extensions of ZF that make the claim true?
2. If the claim can be proven with some other axioms, what does the proof look like?
3. Are there any known axioms that make the claim false?
4. Does the claim have any interesting (set-theoretical) corollaries?

This claim was also considered in this earlier post, but from a different point of view.

A sketchy proof of the claim:

$\newcommand{\Q}{\mathbb Q}$ The groups $G_n$ are naturally vector spaces over $\Q$. It is easy to check that the claim is equivalent with $G_n$ and $G_m$ being isomorphic as vector spaces over $\Q$.

All vector spaces have a Hamel basis; let the basis of $G_n$ be $B_n$. Thus $G_n=\Q^{(B_n)}$ (brackets meaning only finitely many nonzero components). But $|\Q^{(B_n)}|=|B_n|$ since the basis is infinite, so $|B_n|=|G_n|=|\mathbb R|$.

There is a bijection between the bases of the vector spaces $G_n$ and $G_m$ since they have the same cardinality. This gives a linear bijection between them.

Amongst these are "every set is Lebesgue measurable" and "every set has the Baire property". Both are consistent with $\sf ZF+DC$, so we know that dependent choice and countable choice don't suffice for this proof.
On the other hand, if there is a Hamel basis for $\Bbb R$ over $\Bbb Q$, then the usual "choice-y" proof works out just fine. I don't think it is known whether or not the existence of such isomorphisms imply there is a Hamel basis. If you are looking for more set theoretical statements, then the assumption $2^{\aleph_0}=\aleph_1$ (or even just that $\Bbb R$ can be well-ordered) also suffice for all the choice-using arguments to go through without a problem.