Which vector spaces are algebraic dual spaces?

(A note on notation: I will use simple vertical bars to denote the cardinality of a set. There isn't anything else involved to confuse it with.)

With just the algebraic structure, the only thing that matters is the cardinality of the basis. Let a basis of $U$ be $S$; the dual space $U^*$ can be realized as the space of functions from $S$ to the base field $F$. The cardinality there, $|F|^{|S|}$, comes out as whichever of $|F|$ and $2^{|S|}$ is larger.

Now, consider a vector space $V$ with an infinite basis $T$. This space can be realized as the space of finitely supported functions from $T$ to $F$. The cardinality here is the product of cardinalities $|T|\cdot |F|$, which is just whichever one is larger.

So then, if $|F|$ is small - say, if we're working with one of the prime fields $\mathbb{Z}/p$ or $\mathbb{Q}$, the cardinality of an infinite-dimensional vector space will be that of its basis, and the cardinality of a dual will be that of the basis' power set. The possible infinite cardinals that can be the dimension of a dual space are simply the power sets of infinite cardinals.

Now, what about larger fields? If we extend to a larger field while keeping the same basis, that's the same dimension for the vector space. What about its dual? Can we keep the same basis there?

Let $F$ be an extension of the prime field $K$, and let $B$ be a basis for $F$ over $K$. Suppose $U$ has basis $S$ over $K$, and $U'$ is the extension with basis $S$ over $F$. Then let $V=U^*$ and $V'=(U')^*$, with $T$ a basis for $V$ (over $K$).
Is $T$ linearly independent over $F$ in $V'$? Well, we can break down any linear dependence relation into components, one for each element of $B$ - and since $B$ is linearly independent over $K$, each component must give a linear dependence relation $\sum_i a_i (b_j t_i)$ over $K$. The elements of $T$ are linearly independent over $K$, so each of these linear dependence relations is trivial. Sum over $B$, and all the coefficients $\sum_j a_ib_j$ are trivial in $F$, so $T$ is linearly independent over $F$ as well. That's one direction - the dimension of $V'$ over $F$ is at least as large as that of $V$ over $K$, previously calculated as $2^{|S|}$.
Does $T$ span $V'$? Unfortunately, not necessarily. If we take a finite linear combination over $F$ of elements of $T$, the $K$-span of the values these functions take is contained in the $K$-span of the coefficients of the linear combination - which is finite-dimensional. If $F$ is large enough, there will be elements of $V'$ with function values that have an infinite-dimensional $K$-span.

Now, all of that isn't too much trouble in most cases. As long as $|F|\le 2^{|S|}$, we still have that upper bound to lock down the dimension of the dual space as $2^{|S|}$. This covers almost all examples, including all spaces over such fields as $\mathbb{R}$ or $\mathbb{C}$. Really big fields - I'll keep thinking.
[Added in edit] As noted in linked material, it is a known theorem that in those large field cases, the dimension is as large as possible - the full $|F|^{|S|}$. Such a number is always the cardinality of a power set, so no new possibilities for the dimension of a dual set are introduced this way.