Assume that $\det\Sigma\ne0$. Then the random matrix $X$ is of rank $m$ almost surely (a.s.). So, a.s. the Moore--Penrose inverse of $X$ is $X^+=X^\top(XX^\top)^{-1}$ and hence $$X^+X=X^\top(XX^\top)^{-1}X.$$ Letting
It appears that $Z$$EX^+X=EX^\top(XX^\top)^{-1}X$ cannot be expressed in closed form, even in the matrix with rowsfully specified case when $z_i:=\Sigma^{-1/2}x_i$$m=2$, $n=3$, and $\Sigma=\left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$.
Indeed, in this case $\Sigma=A^\top A$ for $A:=\left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$. So, we see thatfor the rows $x_i$ of $X$ we can write $x_i=z_i A$, where the $z_i$'s are iid rows of iid standard normal random variables $N(0,I_n)$, and $$X^+X=X^\top(XX^\top)^{-1}X=Z^\top(ZZ^\top)^{-1}Z=Z^+Z.$$ So$z_{i,j}$.
In the image of a Mathematica notebook below, the problem reduces toexpression of even the case when $\Sigma=I_n$ -$(1,1)$- whichentry (P11) of the matrix $P:=X^+X$ in terms of the $z_{i,j}$'s looks very formidable, as you notedand Mathematica cannot do anything for the expectation of P11, is easy.leaving it unevaluated after working on it for more than an hour (click on the image to enlarge it):
Thus, the answer is
$$EX^+X=\frac mn\,I_n,$$
as long as $\det\Sigma\ne0$.