Let $k$ be a field, and $\widehat{k}$ a field extension of $k$. Then $\widehat{k}$ can be interpreted as a vector space over $k$.
Is some converse also true ?
That is: given $k$, is there a criterion for a vector space $V$ over $k$ so that $V$ arises from a field extension as above ?
I am not restricting to finite-dimensional spaces (so the Axiom of Choice might come into play).
You are asking if every vector space $V$ over $k$ can be endowed with the structure of a field extension of $k$. The only invariant of a vector space is its dimension, so the question can be rephrased as:
As often happens for this type of cardinality question, it is actually the finite case that is the interesting one. If $[l:k]$ is finite, then $l/k$ is algebraic. If we suppose that $k$ is perfect (e.g. when $k$ has characteristic zero) then by the Primitive Element Theorem we then have $l = k[\alpha]$ for some $\alpha \in l$, so that there is some monic irreducible polynomial $P \in k[t]$ such that $l \cong k[t]/(P)$. In other words, over a perfect field the question of which finite numbers arise as field extensions is equivalent to the question of what are the degrees of monic irreducible polynomials over $k$. It is well known that the answer depends on $k$. For instance, if $k = \mathbb{C}$ or is algebraically closed then the only possible degree is $1$, and if $k = \mathbb{R}$ (or is real-closed) then the only possible degrees are $1$ and $2$. On the other hand, if $k$ is a finite field or $k = \mathbb{Q}$, for instance, then all finite degrees occur.
In contrast, I claim that for any field $k$ and any infinite cardinal $\kappa$, there is a field extension $l/k$ with $\operatorname{dim}_k l = \kappa$. Indeed, let $\{t_i\}_{i \in \kappa}$ be a set of indeterminates indexed by $\kappa$ and take $l = k(t_i)$ to be the corresponding rational function field (the fraction field of the ring of polynomials in these indeterminates).