What is $\dim(V^V)$? Where $V$ is a vector space and $V^V=\left\{ f\mid f\colon V\rightarrow V \right\}$. 

I proved that $V^V$ is a vector space so $\dim\left(V^V\right)$ is well defined.
If $\dim V=0$, then one readily finds that $V^V$ is a singleton set, hence $\dim(V^V)=0$, too.

If $V$ is a vector space over a finite field $\Bbb F_q$ and of finite dimension $n$, then $V$ is a set of $q^n$ elements. Then one quickly finds that $V^V\cong V^{q^n}$ is of dimension $n^{q^n}$.

If $V$ is an infinite set (that is, either of infnite dimension over an arbitrary field, or of any positive dimension over in infinite field), then certainly $\dim(V^V)$ is infinite. If needed, you could still make more precise statements about the exact - infinite - cardinality, but I suppose yo do not want to go down that rabbit hole ...
If V has finite dimension n then $V^V$ has dimension $n^n$.  If V has countable dimension then $V^V$ has countable dimension.  If V has uncountable dimension then $V^V$ has uncountable dimension.