Let $E/\mathbb{Q}$ be an elliptic curve and let $p$ be a prime. Then there is an action of the absolute Galois group of $\mathbb{Q}$ on $E[p]$ that factors through a finite quotient.
Does any finite quotient of the absolute Galois group of $\mathbb{Q}$ arise this way for some $E$ and $p$? What if we consider abelian varieties instead of elliptic curves?
For elliptic curves, the answer is no, as Chris Wuthrich and Aron Fehm point out in the comments. In fact I think every extension with Galois group a finite simple group not of the form $PSL_2(\mathbb F_p)$ for any $p$ cannot arise this way.
For abelian varieties, the answer is positive. In fact it's sufficient to take just the $2$-torsion, and even just the $2$-torsion of Jacobians of hyperelliptic curves. Given an arbitrary extension, take the minimal polynomial $f$ of a generator $\alpha$ of that extension, form the hyperelliptic curve with equation $y^2 = f(x)$, and take its Jacobian. (If the degree of $f$ is at most $4$, you will need to add some linear factors as well to make it hyper-.) The Galois group of the $2$-torsion of the hyperilliptic curve will be a subgroup of $S_n$, generated by the Galois action on the roots of this polynomial (plus the point at $\infty$, if the degree is odd) and will therefore include the Galois group of your field.
