Given a field $\mathbb F$, is there a smallest field $\mathbb G\supseteq\mathbb F$ where every element in $\mathbb G$ has an $n$th root for all $n$?

Well, from my thrashing-about in the comments, it’s probably clear that I haven’t fully understood the depth of this problem. But let me make some remarks anyhow, first about the very special case $\mathbf F=\Bbb F_2(x)$. It happens, and some highly-experienced people seem not to know this, that when $\mathbf F$ is of transcendence degree one over a perfect field of characteristic $p$ and not already perfect, there is exactly one radicial (=purely inseparable, that’s French radiciel) extension of each possible degree $p^m$. I think this should clarify your thinking on a square-root-closed extension of $\Bbb F_2(x)$.

Second, let me just point out the difficulty of describing any construction for your field $\mathbf F=\Bbb F_2(x)$: for $d$ odd, the extensions $\mathbf F(\sqrt[d]x\,)$ and $\mathbf F(\sqrt[d]{x+1}\,)$ have nothing to do with each other: their intersection is the ground field $\mathbf F$. Stick any $\Bbb F_2$-irreducible polynomial under the radical sign and get another totally unrelated extension. So conceptually anyway, this is getting to be a mess; you need to worry about rational expressions, too.

I guess that you were thinking of specifying at the outset that in every case, the $n$-th root of $1$ that you choose is $1$ itself. Even if you do that, it’s not clear to me that in your choosing lots of unspecified $n$-th roots of other elements, you may inadvertently induce the presence of other roots of unity than $1$ itself. This would actually simplify matters, but I do think you’re making things hard for yourself. It seems to me that if you agree at the outset that all roots of unity should be added (this would make the constant field $\overline{\Bbb F_2}\,$, algebraically closed), then the existence of your field is now easily seen, though an explicit construction would remain still something of a problem.

Here is a good benchmark. If $\Bbb F$ is a field, and $\Bbb K$ is a field extension such that:

1. $\Bbb{K/F}$ is infinite dimensional, and
2. for every irreducible $p\in\Bbb F[x]$, if $p$ has at least two roots in $\Bbb K$, then there is an automorphism of $\Bbb K$ (fixing $\Bbb F$) which is a complete derangement of the roots of $p$ in $\Bbb K$.

In this case, there is a model of $\sf ZF$ in which there is a field which is "morally isomorphic to $\Bbb K$, but not internally isomorphic to it". That is to say, we add a new copy of $\Bbb K$, but we remove the isomorphism, and indeed any bijection, while preserving the field structure, and every field extension of $\Bbb F$ embedding to both will be finite dimensional.

It's not hard to see that your "smallest field" will satisfy these properties, or at the very least, we can find such a field which will ensure no "smallest" exists.