Take $\pi^N$ with $AW(\pi^N, \pi) \leq \frac{1}{N}$, where we denote by $\mu^N$ and $\nu^N$ the marginals of $\pi^N$.
Note that by the backward induction for $AW$ (cf. here), it holds $$ AW(\pi, \hat\pi^N) = \inf_{\kappa_1 \in \Pi(\mu, \mu^N)} d_{X_1}(x_1, y_1) + W_1(\pi_{x_1}, \pi^N_{y_1}) \kappa_1(dx_1, dy_1), $$ and thus we can choose $\kappa_1^N \in \Pi(\mu, \mu^N)$ such that $$ \int W_1(\pi_{x_1}, \pi^N_{y_1}) \kappa_1^N(dx_1, dy_1) \leq \frac{1}{N}, $$ and further clearly the second marginal converges, i.e., $W_1(\nu^N, \nu) \leq \frac{1}{N}$.
By applying twice the triangle inequality, we get \begin{align} \int W_1(\pi_{x_1}, \nu) \mu(dx_1) - \int W_1(\pi^N_{y_1}, \nu^N) \mu^N(dy_1) &= \int W_1(\pi_{x_1}, \nu) - W_1(\pi^N_{y_1}, \nu^N) \kappa_1(dx_1, dy_1) \\ &\leq \int W_1(\nu, \nu^N) + W_1(\pi_{x_1}, \pi_{y_1}^N) \kappa_1^N(dx_1, dy_1) \\ &\leq \frac{2}{N} \end{align} and vice versa.
Since the denominator of $W(\pi)$ should be strictly larger than zero (and converges as well since $W_1(\nu^N, \nu)$ goes to zero), we get that the estimator is consistent.