For Hadamard product, you have (source):
$$
\text{rank}(A\odot X) \leq \text{rank}(A) \text{rank}(X)
$$
This is as tight as you can go even with knowing $\text{rank}(X)$.
Without knowing anything about $X$, there is certainly nothing you can say. (See the examples that nicoguaro and Wolfgang Bangerth brought in the comments section).
Now, imagine you know $\text{rank}(X)=r$, and $\text{rank}(X)\geq1$ (I think the without loss of generality clause is applicable here).
- If $\text{rank}(A)=d$, then the inequality above is totally useless, as it would not bound anything: $d\cdot r \geq d$, which is the same conclusion you would get from just looking at $A\odot X \in \mathbb{F}^{d\times n}, d<n$.
- If $\text{rank}(A)=r<d$, depending on what you know about $X$ can give you something, but still not too exciting.
Hadamard products are tricky, and a lot of things one expects from them (based on prior knowledge of the regular matrix-matrix product) do not hold.