This is not quite an answer, but not quite a comment either.

We can at least show that $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$ is infinite (where $X$ is the quintic Fermat curve). There are in fact (at least) two ways of showing this.

- $X$ has an automorphism $\tau$ of order 3 defined over $\mathbb Q$ (rotate the projective coordinates). The quotient $X/\langle \tau \rangle$ is a hyperelliptic curve $Y$. As I pointed out in a related thread, $Y({\mathbb Q}^{\text{ab}})$ is infinite for every hyperelliptic curve $Y$. Now the preimages on $X$ of any ${\mathbb Q}^{\text{ab}}$-point on $Y$ are defined over a cyclic degree-3 extension of ${\mathbb Q}^{\text{ab}}$, so are in $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$.
- Fix any $y \in {\mathbb Q}$, then the points $(x:y:1) \in X$ have $x$ in ${\mathbb Q}(\mu_5, \sqrt[5]{1-y^5}) \subset ({\mathbb Q}^{\text{ab}})^{\text{ab}}$.

The second construction obviously generalizes to arbitrary Fermat curves.
In my originial post, I claimed that the first one does, too. But this seems to be wrong: the quotient of the degree $n$ Fermat curve $x^n + y^n + z^n = 0$ by the obvious $S_3$-action has positive genus as soon as $n \ge 6$, so the quotient by the cyclic group is no longer obviously hyperelliptic. (It is true, however, that this curve maps to the hyperelliptic curve $y^2 = x^n + \frac{1}{4}$, but this is via the quotient w.r.t. the action of $\mu_n$ such that $\zeta$ sends $(x:y:z)$ to $(\zeta x : \zeta^{-1} y : z)$. This would lead to a weaker conclusion, replacing the double by the triple abelian closure of $\mathbb Q$. I had mixed up the two quotients.)

Note that these constructions use the fact that $X$ has (many) nontrivial automorphisms. So perhaps another interesting question is the following.

*Let $C$ be a nice (smooth, projective, geometrically irreducible) curve over $\mathbb Q$. Can we at least show that $C({\mathbb Q}^{\text{sol}})$ is infinite?*

Here ${\mathbb Q}^{\text{sol}}$ denotes the union of all finite Galois extensions of $\mathbb Q$ with solvable Galois group.

(Originally the question was formulated for (sufficiently generic) curves of genus 3. However, as René pointed out, we find lots of quartic points by intersecting the plane quartic $C$ with rational lines, and all quartic fields are contained in ${\mathbb Q}^{\text{sol}}$.)

EDIT: The question whether solvable points exist has been studied, see for example a recent preprint by Trevor Wooley and the references given there, in particular this paper by Ambrus Pál.

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