Usual approaches to the inverse Galois problem start with realizations of a group $G$ over a larger field, and then try to specialize to ${\Bbb Q}$.

One could also start by building suitable objects over ${\Bbb Q}$ and then trying to locate among them a field.

Writing $n=|G|$, we could start with $A=<G>_{\Bbb Q}$, the $n$-dimensional ${\Bbb Q}$-vector space with basis $G$. Then equipping $A$ with a product $\ast$ (not to be confused with the group operation of $G$) making it a (not necessarily associative) algebra just means choosing structure constants $c_{g,h,j}\in {\Bbb Q}$ so that $$ g \ast h = \sum_{j\in G} c_{g,h,j} j\ .$$

Next we want the natural action of $G$ on the basis of $A$ to preserve $\ast$, so for $k\in G$, $$ kg \ast kh = \sum_{j\in G} c_{g,h,j} kj = \sum_{j\in G} c_{g,h,k^{-1}j} j\ .$$ This means $c_{kg,kh,j} = c_{g,h,k^{-1}j}$. Making $\ast$ commutative also forces $c_{g,h,k}=c_{h,g,k}$.

With no loss of generality, we can make the group identity $e\in G$ serve also as the identity for $\ast$, with $c_{e,h,j}=\delta_{h,j}$.

By linearity, imposing associativity on $A$ merely requires $(g\ast h)\ast j = g\ast (h\ast j)$ for basis elements. Since the action of $G$ preserves $\ast$, we can even fix $g$, say as $e$, the identity of $G$, a further reduction. So associativity imposes a family of quadratic relations on the structure constants. In any event, we get an affine variety $V_G$ that parametrizes commutative, associative algebras with a $G$-action that looks like the regular representation of $G$. (Is anything known in general about these varieties?)

It seems to me that the real difficulty here lies in requiring $\ast$ to form a field. I suppose it suffices to make $A$ a domain, and thus make multiplication by each element $\alpha=\sum_{g\in G} a_g g\in A$ injective. For a fixed $\alpha$ this comes down the the non-vanishing of a determinant, but then we must quantify over all non-zero $\alpha\in A$, so getting an algebraic condition on the structure constants involves some elimination of quantifiers or classical elimination theory.

But the difficulty here reminds me of the classical story about classifying division algebras over ${\Bbb R}$, with the related story about the existence of orthogonal families of vector fields on spheres and links to $K$-theory. I'm not expert in those matters, but I wonder whether analogous ideas might have some bearing here.

Usual approaches to the inverse Galois problem start with realizations of a group $G$ over a larger field, and then try to specialize to ${\Bbb Q}$.

One could also start by building suitable objects over ${\Bbb Q}$ and then trying to locate among them a field.

Writing $n=|G|$, we could start with $A=\langle G\rangle_{\Bbb Q}$, the $n$-dimensional ${\Bbb Q}$-vector space with basis $G$. Then equipping $A$ with a product $\ast$ (not to be confused with the group operation of $G$) making it a (not necessarily associative) algebra just means choosing structure constants $c_{g,h,j}\in {\Bbb Q}$ so that $$ g \ast h = \sum_{j\in G} c_{g,h,j} j\ .$$

Next we want the natural action of $G$ on the basis of $A$ to preserve $\ast$, so for $k\in G$, $$ kg \ast kh = \sum_{j\in G} c_{g,h,j} kj = \sum_{j\in G} c_{g,h,k^{-1}j} j\ .$$ This means $c_{kg,kh,j} = c_{g,h,k^{-1}j}$. Making $\ast$ commutative also forces $c_{g,h,k}=c_{h,g,k}$.

With no loss of generality, we can make the group identity $e\in G$ serve also as the identity for $\ast$, with $c_{e,h,j}=\delta_{h,j}$.

By linearity, imposing associativity on $A$ merely requires $(g\ast h)\ast j = g\ast (h\ast j)$ for basis elements. Since the action of $G$ preserves $\ast$, we can even fix $g$, say as $e$, the identity of $G$, a further reduction. So associativity imposes a family of quadratic relations on the structure constants. In any event, we get an affine variety $V_G$ that parametrizes commutative, associative algebras with a $G$-action that looks like the regular representation of $G$. (Is anything known in general about these varieties?)

It seems to me that the real difficulty here lies in requiring $\ast$ to form a field. I suppose it suffices to make $A$ a domain, and thus make multiplication by each element $\alpha=\sum_{g\in G} a_g g\in A$ injective. For a fixed $\alpha$ this comes down the the non-vanishing of a determinant, but then we must quantify over all non-zero $\alpha\in A$, so getting an algebraic condition on the structure constants involves some elimination of quantifiers or classical elimination theory.

But the difficulty here reminds me of the classical story about classifying division algebras over ${\Bbb R}$, with the related story about the existence of orthogonal families of vector fields on spheres and links to $K$-theory. I'm not expert in those matters, but I wonder whether analogous ideas might have some bearing here.