We claim that the maximal length of a walk is $3 + 4\sqrt{2}$. An example of such a walk is $A \to A' \to B \to D \to D' \to C \to B' \to C'$. Since the cube has $8$ vertices, the ant's walk consists of at most $7$ steps, each of length either $1$ or $\sqrt{2}$.

The length of a $6$ steps walk is at most $6\sqrt{2} < 3 + 4\sqrt{2}$, hence the maximal walk must have exactly $7$ steps.

We claim that the number of "diagonal" steps is at most $4$. For this purpose, let us color black the vertices $A, C, B', D'$, and white the other four. Observe that a "diagonal" step changes the color, while an "edge" step changes it. Since $A$ and $C'$ have different colors, we deduce that the walk must contain an odd number of "edge" steps. By inspection, we see that a walk with one "edge" step and $6$ "diagonal" ones self-intersects, thus proving the claim.