Non-algebraically closed field in which every polynomial of degree $<n$ has a root

I’ll give you a hint, and not an answer. Best route to understanding here is to use Galois Theory. The total Galois group of a finite field $k$, i.e. the group of the algebraic closure over $k$, is $\hat{\mathbb Z}$, the profinite completion of the integers. It’s topologically generated by the single automorphism, the Frobenius of $k$. To understand $\hat{\mathbb Z}$, use Chinese Remainder Theorem, and you see that it’s the direct product of all the groups ${\mathbb Z}_p$, with $p$ running through all the primes. You take it from there.

If you start with a field of size q, and adjoin a root of any (all) irreducible quadratic, you get a field of size $q^2$.

Start with $q=2$, and you get 2, 4, 16, 256, etc. None of these fields contains a root of an irreducible cubic over the original field (with $q=2$, that would require a field whose size was a power of 8).

In other words, you don't get the algebraic closure, since for any prime r bigger than n, you don't get the roots of any irreducible polynomials of degree r.

As Lubin mentions, this is equivalent to taking a Sylow pro-2-subgroup of the Galois group of the algebraic closure, and I guess in general you want a Hall pro-n-subgroup of the Galois group, but I prefer just thinking about repeatedly squaring a number.