Do $K(A)$ and $K(\hat A)$ have isomorphic absolute Galois groups?

@JasonStarr Thank you, I didn't find a reference for the Greenberg spaces, do you mean Weil restriction. My motivation is edited as above.

Now I understand, thank you. Besides, I find a proof using Fitting lemma math.stackexchange.com/questions/1298453/…

@WillSawin So you finish by induction on dimension? Could you provide some details on "Sufficient to show....This is equivalent....exceptional divisor. "? As for any projective variety X we can produce an affine cone whose number of rational points is related to number of X in a simple way, so isn't the vanishing result a restriction on X?

@WillSawin Thanks, how do you compute its cohomology in char 0 ? As base change fails in general non-smooth case, betti number of special fiber and generic fiber might be different.

@WillSawin Thank you, I am not familiar with the positive characterestic case. I am mostly interested in Q2 and compute the case $n=2$.

Yes, I mean the permutation representation. In general you could assume $A$ is symmetric or regard it as a quadratic form.