Is my field algebraically closed? 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.) 
 A: $\def\QQ{\mathbb{Q}}$Building on YCor's construction, your field is $\overline{\mathbb{Q}}$. Let $L$ be any finite Galois extension of $\QQ$ with Galois group $G$; I will show that $L$ is contained in your field. 
We can find an element $x$ in $L$ so that, for any nonempty subset $S$ of $G$, the product $\prod_{\sigma \in S} x^{\sigma}$ is not square. For example, choose $x$ to generate a principal prime ideal of $L$ lying over a prime of $\mathbb{Q}$ which is completely split in $L$. Let $M$ be the result of adjoining to $L$ all square roots $\sqrt{x^{\sigma}}$, for all $\sigma \in G$. So $Gal(M/\mathbb{Q}) \cong G \ltimes (\mathbb{Z}/2)^G$, where the first $G$ acts by permuting the factors in the second $G$. 
Let $F \subset M$ be the fixed field of $G \ltimes \{ 0 \}$. This group has index $2^{|G|}$ in $Gal(M/\mathbb{Q})$, so $[F:\QQ]=2^{|G|}$. Let $f$ be the minimal polynomial of a primitive element of $F$. So $f$ splits in your field. Thus, your field contains $F$ and all Galois conjugates of $F$. But the intersection $\bigcap_{h \in G \ltimes (\mathbb{Z}/2)^G} h (G \ltimes \{ 0 \}) h^{-1}$ is trivial, so the collection of Galois conjugates of $F$ generates $M$. We have shown that your field contains $M$, and hence contains $L$.
