If $A^3=0$, then the minimal polynomial divides $t^3$, and that means that the characteristic polynomial must be $t^5$ (alternatively, if $\lambda$ is an eigenvalue of $A$, then $\lambda^3$ is an eigenvalue of $A^3$, so $\lambda^3=0$, hence $\lambda=0$).
Now remember that the highest power of $t-\lambda$ that divides the minimal polynomial gives you the size of the largest Jordan block associated to $\lambda$ in the Jordan canonical form. So, for example, if the minimal polynomial were $t^4$, that means that there must be at least one block of size $4$, and no blocks of larger size. For a $5\times 5$ matrix, this would mean that the Jordan form must consist of a single $4\times 4$ Jordan block associated to $0$, and a $1\times 1$ block associated to $0$ (since that is all that is left).
Now consider the three possibilities for the minimal polynomial and what that tells you. Enumerating the possible Jordan forms from that information is fairly straightforward.
As a way to check your work, there should be five different possible Jordan canonical forms.