Solution:

Note that $\angle A C D = \angle A C B + \angle B C D = 60^{\circ} + 90^{\circ} = 150^{\circ}$. In a dodecagon, each interior angle is $180^{\circ} \cdot \frac{12-2}{12} = 150^{\circ}$, meaning that $\angle F E D = \angle D O N = 150^{\circ}$. Since $E F = F D = 1$ and $D O = O N = 1$ (just like how $A C = C D = 1$), then we have that $\triangle A C D \cong \triangle D O N \cong \triangle F E D$, and because the triangles are isosceles, then $A D = D F = F N$, so $D$ is the circumcenter of $\triangle A F N$.

Now, applying the Law of Cosines gets that $A D^{2} = 2 + \sqrt{3}$, so $A D^{2} \pi = (2 + \sqrt{3}) \pi$.