It may be more helpful to think about the relationship you need to show in terms of the Hermitian form $\sum \limits_{m,n=0}^{d-1} U_{m,n} U_{m,n}^\ast = d$, which is a necessary condition for the defining unitary relationship $UU^\dagger = I$.

To see this explicitly, consider the four equations implicit in the $d=2$ case: $$UU^\dagger=\begin{bmatrix} U_{1,1} & U_{1,2} \\ U_{2,1} & U_{2,2} \end{bmatrix} \begin{bmatrix} U_{1,1}^\ast & U_{2,1}^\ast \\ U_{1,2}^\ast & U_{2,2}^\ast \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$
The diagonal terms are what you need for the desired relationship:
$$(U_{1,1} U_{1,1}^\ast + U_{1,2} U_{1,2}^\ast) + (U_{2,1} U_{2,1}^\ast + U_{2,2} U_{2,2}^\ast) = 1 + 1 = 2.$$
It should not be hard to convince yourself that this generalizes to any $d$.