Nice question, Jonas!
Yes, $M[j''\text{Ord}]=V$. To see this, fix any set $a\in V$. We may by standard coding methods code $a$ with a subset $A\subset\kappa$ for some cardinal $\kappa$. Thus, $a$ is determined by $A$. But also, from $j(A)$ and $j''\kappa$ we can reconstruct $j''A$ and hence $A$ and hence $a$. But $j(A)\in M$, and so from objects in $M$ and $j''\kappa$, we can construct $a$. So $M[j''\text{Ord}]=V$. QED