Does the dualizing process on vector spaces necessarily terminate? It's well-known (assuming the axiom of choice) that the inclusion $\ell^1 \subset (\ell^1)^{**}$ is proper as a simple corollary of the Hahn-Banach theorem. But is this the end of the dualizing process; e.g., does $(\ell^1)^{**} = (\ell^1)^{****}$, or do we continue to get more unwieldy vector spaces over time? If this process does not terminate, is there a non-trivial (i.e. not 'reflexive') condition on vector spaces $V$ such that, for some large enough $n$, $V^{2n*}=V^{(2n+2)*}$?
On a similar topic, if we deny the axiom of choice, we know that $(\ell^1)^{**}=\ell^1$ (source). In this case, do we also have $V^{**}=V$ for all $V$?
 A: Interestingly, there are exactly two possibilities: a Banach space is either reflexive or its chain of double duals never terminates. This only uses a couple of facts: $(1)$ a Banach space is closed in its dual and $(2)$ a closed subspace of a reflexive vector space is reflexive. $(1)$ is a fact about any complete subspace of a metric space. 
Here's  $(2)$, suppose $X\subset B$ is closed and $B=B^{**}$. Then any element $\xi$ of $X^{**}$ corresponds to an element $\xi_B$ of $B^{**}$ acting on $L\in B^*$ by restriction, $\xi_B(L)=\xi(L|_X)$. On the other hand since $B$ is reflexive $\xi$ corresponds uniquely to $x\in B$, that is, $\xi_B(L)=\xi(L|_X)=L(x)$. This shows that if $L$ vanishes on $X$, $L(x)=0,$ so since $X$ is closed $x\in X$.
So, suppose $X$ is a Banach spaces and its chain of double duals eventually stabilizes, say the $n$th double dual is isomorphic to the $n+1$st. Then the $n$th double dual is reflexive, and being a closed subspace, so is $X$ itself.

I was going to avoid commenting on the set theoretic part of your question, since I don't know much about alternatives to the axiom of choice. But having just checked your link the answer, which is no, is actually there: in Zermelo-Frankel set theory with dependent choice and the axiom that all sets of real numbers have the property of Baire, the double dual of $(\ell^\infty/c_0)^{**}=0$. But it certainly doesn't require the axiom of choice to show there are bounded sequences that don't converge to zero! So $\ell^\infty/c_0\neq 0$ even in that system, and in the absence of choice a Banach space needn't even embed into its double dual.
A: Allow me to comment on the axiom of choice related question.
First of all, it is far from true that " if we deny the axiom of choice, we know that $(\ell^1)^{**}=\ell^1$". That is consistent with the failure of the axiom of choice, but it is certainly not "true if we deny the axiom of choice". We may have a situation where $\Bbb R$ can be well-ordered and still the axiom of choice fails in the most acute ways; or we can have that even $\Bbb R$ cannot be well-ordered, but the Hahn-Banach theorem holds, in which case $\ell^1\subsetneq(\ell^1)^{**}$.
And to the second point, it is never the case that for every vector space $V^{**}=V$. Note that given any ordinal $\alpha$, we can look at the direct sum of $\alpha$ copies of $\Bbb R$, and endow this with a norm (take the maximal absolute value of non-zero coefficient of a vector in a given basis, for example). The dual will be strictly larger, as one can show via diagonalization. Note that this is not necessarily a Banach space, though.
If we do insist on Banach spaces, then we may observe the following:


*

*If $(\ell^1)^{**}=\ell^1$ then $(\ell^\infty/c_0)^*=\{0\}$, while $\ell^\infty/c_0\neq\{0\}$, and clearly the dual of $\{0\}$ is just $\{0\}$.

*Otherwise, $\ell^1\neq(\ell^1)^{**}$ and we are done.


In general, if one allows to consider the algebraic dual (since continuous dual requires a topology) then it can also be the case where an infinite dimensional vector space is isomorphic to its dual (e.g. in the case where all functionals are automatically continuous, the topological dual and algebraic dual are the same for Banach spaces); or we can have vector spaces whose dual is zero. In fact we can have vector spaces which cannot be topologized in a nontrivial sense.
