Let $\mu$ be the uniform measure on the cube $Q = [-1,1]^n$, and $\nu$ be the uniform measure on the surface $$ V = \{(x_1,\dots,x_n)\in Q \mid \sum x_i = 0\}. $$ I am curious about Wasserstein distance between $\mu$ and $\nu$, using the $l^\infty$ distance on $\mathbb{R}^n$.
That is, I would like a bound on $$ \sup_{\|f\|_{Lip}=1} |\mathbb{E}_\mu f - \mathbb{E}_\nu f|. $$ I expect that the distance is on the order of $n^{-1/2}$ based on the following two calculations. First, one guess is that the function $S(x)=\frac{1}{n}|\sum x_i|$ would be close to optimizing the above supremum, and $$ |\mathbb{E}_\mu S - \mathbb{E}_\nu S| \leq Cn^{-1/2}. $$
The second heuristic follows from the fact that if $x\in Q$ is sampled according to $\mu$, then with high probability $|\sum x_i| \leq Cn^{1/2}$, so by nudging each of the coefficients by an amount $n^{-1/2}$ it should be possible to land on the plane. This gives a transport map from $\mu$ to a measure supported on $V$ that has cost roughly $n^{-1/2}$. The problem is that the transported measure is not the same as $\nu$, so this does not constitute a proof.