Let $g \sim N(0, \frac 1 d I)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$, so $(x_1, \|g\|x_1)$ is a coupling between the marginal distribution of $x_1$ and $N(0, 1/d)$.
The norm $\|g\|$ is sharply concentrated around $1$, with fluctuations of order $1/\sqrt{d}$, so the Wasserstein distance is bounded above by $$\mathbb E|x_1 - \|g\|x_1| = (\mathbb E|1 - \|g\||) \mathbb E|x_1| = O(d^{-1}), $$ where we have used the fact that $\mathbb E|x_1| \le \sqrt{\mathbb E|x_1|^2} = \sqrt{1/d}$.
