Is every $GL_2(\mathbb{Z}/n\mathbb{Z})$-extension contained in some elliptic curve's torsion field?

I have two things to add to this discussion.

$\bullet$ For $n = 2$ and $n = 3$, every Galois extension of $\mathbb{Q}$ with Galois group ${\rm GL}_{2}(\mathbb{Z}/n\mathbb{Z})$ does arise from an elliptic curve (by a result of Shepard-Barron and Taylor from 1997 - see the reference in the paper of Dieulefait linked to below.) For $n = 4$, this is not true (see this paper).

$\bullet$ For $p \geq 7$ prime, it is known that not every Galois representation $\rho : G_{\mathbb{Q}} \to GL_{2}(\mathbb{F}_{p})$ arises from an elliptic curve, even if we restrict $\rho$ to have cyclotomic determinant. This is shown in a paper of Dieulefait for $p \geq 7$, because one can construct Galois representations using modular forms of different weights. This somehow doesn't seem quite the same as starting with a field $K/\mathbb{Q}$ with $Gal(K/\mathbb{Q}) \cong GL_{2}(\mathbb{F}_{p})$. (Any two isomorphisms between $Gal(K/\mathbb{Q})$ and $GL_{2}(\mathbb{F}_{p})$ differ by an automorphism of $GL_{2}(\mathbb{F}_{p})$ and I haven't been able to find a convenient reference for what ${\rm Aut}(GL_{2}(\mathbb{F}_{p}))$ is.)

This is likely a very hard problem once $n \geq 7$.

Such $E$ are parametrized by a twist of the modular curve ${\rm X}(n)$; this curve is rational for $n \leq 5$ (so for those $n$ one can use the Hasse principle), but of genus at least $3$ for $n\geq 7$. Thus if $n \geq 7$ there are at most finitely many $E$ (Mordell-Faltings) but we do not have a general technique for deciding whether one exists. This was the main difficulty that Poonen, Schaefer, and Stoll had to overcome for several special choices of $K$ in their paper

Bjorn Poonen, Edward F. Schaefer, and Michael Stoll: Twists of $X(7)$ and primitive solutions to $x^2+y^3=z^7$, Duke Math. J. 137 #1 (2007), 103-158 (arXiv:math/0508174).

(For $n=6$ the modular curve ${\rm X}(6)$ is isomorphic with the CM elliptic curve $Y^2 = X^3 + 1$, and the problem reduces to deciding the existence of rational points on a principal homogeneous space for a twist $Y^2 = X^3 + d$ of this curve; such questions are often tractable in practice, though even here no general algorithm is known.)

This is mostly adding to Vesselin Dimitrov's comment.

If there exists a regular extension of $\mathbb{Q}(t)$ with Galois group $G := GL_2(\mathbb{Z}/n)$ (ie, $\mathbb{Q}$ is algebraically closed inside the extension), then we can construct a $GL_2(\mathbb{Z}/n)$-Galois extension of $\mathbb{Q}$ which does not contain $\mu_n$, and hence cannot contain $K$ (since any such $K$ must be equal to $\mathbb{Q}(E[n])$).

(I believe it's expected, and certainly proved in some cases, that $GL_2(\mathbb{Z}/n)$ occurs as a regular Galois group over $\mathbb{Q}(t)$, though I'm having trouble finding references.)

Specifically, take a regular $G$-Galois extension of $\mathbb{Q}(t)$, which corresponds to a $G$-Galois cover $Y\rightarrow X$, where $X$ is some open subscheme of $\mathbb{P}^1_{\mathbb{Q}}$.

Let $K$ be any number field (for our purposes lets take $K = \mathbb{Q}(\mu_n)$), then by Hilbert irreducibility, $Y_K\rightarrow X_K$ has infinitely many connected fibers. More specifically, the $K$-points of $X_K$ over which the fiber is disconnected is a thin subset of $X_K(K)$ (One key property of thin sets is that subsets of thin sets are also thin). Let "$A$" denote this subset.

Since $Y_K\rightarrow X_K$ is $G$-Galois, these connected fibers are just $G$-Galois field extensions of $\mathbb{Q}$.

Now consider the set "$B$" of $\mathbb{Q}$-points of $X$, viewed as a subset of $X_K(K)$. I.e., given any map $\text{Spec }\mathbb{Q}\rightarrow X$, we can pull back this map by the natural base change map $X_K\rightarrow X$ to get a map $\text{Spec }K\rightarrow X_K$.

This set $B\subset X_K(K)$ is not thin (e.g. Proposition 3.2.1 in Serre's Topics in Galois Theory). Thus, we have two subsets $A,B\subset X_K(K)$, where $A$ is thin, and $B$ is not thin, and so $B$ is not contained in $A$ - In order words, there exist a $K$-point $x_K\in X_K(K)$, which have connected fiber in $Y_K$, which actually come from a $\mathbb{Q}$-point $x\in X(\mathbb{Q})$, but this means that the fiber of $Y\rightarrow X$ above this $\mathbb{Q}$-point $x$ is connected, and so is equal to $\text{Spec }L$ for some $G$-Galois extension $L/\mathbb{Q}$, and furthermore, since the connected fiber of $Y_K\rightarrow X_K$ above $x_K$ is $$\text{Spec }L\times_{\text{Spec }\mathbb{Q}}\text{Spec }K = \text{Spec }L\otimes_{\mathbb{Q}} K$$ this implies that $L\otimes_\mathbb{Q}K$ is a field - in particular $L\cap K = \mathbb{Q}$.

What this says is that if the "regular inverse galois problem" is true for $G$, then for any number field $K$, one can find $G$-Galois extensions of $\mathbb{Q}$ which intersect $K$ trivially.