For a field $L$, let $\widetilde L$ be the splitting field of all irreducible polynomials over $L$ having **prime-power degree**.

Question: Do we have $\widetilde{\mathbf Q}=\overline{\mathbf Q}$?

My money is on "no", because I see no obvious reason why it should be true. If the answer is indeed negative, can one say what degree occurs as the smallest degree of an $f\in \mathbf Q[X]$ which does *not* split in $M$?

In any case, it seems quite difficult (to me)...

*A variation*: We have a chain $$L \subset \widetilde L \subset \widetilde {\widetilde L} \subset \dots$$
Let $\widehat L$ be the limit of this chain. Is it even true that $\widehat{\mathbf Q}=\overline{\mathbf Q}$? Does the chain stabilize? The field $\widehat{\mathbf Q}$ has the strange property of having no finite extensions of prime-power degree. Correspondingly, the Galois group $\text{Gal }(\overline{\mathbf Q}/\widehat{\mathbf Q})$ has the strange property of having no open subgroups of prime-power index...

For $L$ a finite field, it is easy to see that $\widetilde{L}=\overline{L}$. We obviously have $\widetilde{\mathbf R}=\overline{\mathbf R}$. I do not know if $\widetilde{\mathbf Q_p}=\overline{\mathbf Q_p}$, or if $\widehat{\mathbf Q_p}=\overline{\mathbf Q_p}$. (*Edit*: Every finite Galois extension of $\mathbf Q_p$ is solvable, and I believe it follows from this that $\widehat{\mathbf Q_p}=\overline{\mathbf Q_p}$.)

(I have asked this on MSE: please see the discussion there.)