Here is a simple argument showing that, with the above notations, if $U$ is a unitary operator, and $\tilde{U}$ an antiunitary operator in $\bf{C}^n$, one always has $||U-\tilde{U}||\ge \sqrt{2}$.
We note that, by definition of the operator norm we have $||U-\tilde{U}||=\sup_{||z||=1} ||Uz-\tilde{U}z||=\sup_{||z||=1}||z-U^{-1}\tilde{U}z||$. The operator $U^{-1}\tilde{U}$ is antiunitary, hence of the form $z\mapsto U_0\bar{z}$, where $U_0$ is some unitary operator in $\bf{C}^n$. Hence we must show that for any such $U_0$ we have $\sup_{||z||=1}||z-U_0\bar{z}||^2\ge 2$. Now $U_0$ is diagonalizable, and its eigenvalues have unit modulus. Let $u_0$ be any normalized eigenvector, and let $e^{i\phi_0}$ be its eigenvalue. Setting $z=e^{i\alpha}\bar{u}_0$, we have $||z-U_0\bar{z}||^2=||e^{i\alpha}\bar{u}_0-e^{i(\phi_0-\alpha)}u_0||^2=||\bar{u}_0-e^{i(\phi_0-2\alpha)}u_0||^2$ $=2-2\Re\{e^{i(\phi_0-2\alpha)}(\bar{u}_0|u_0)\}$.
It follows that $||U-\tilde{U}||^2\ge \sup_{\alpha\in [0,2\pi]}(2-\Re\{e^{i\alpha}(\bar{u_0}|u_0)\})\ge 2$, Q.E.D.
I suspect one could do better (for example when $n=1$ one finds easily that always $||U-\tilde{U}||=2$), but this is not the point here.