Slightly weaker bound via a more robust method
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Below I present an alternative method with slightly worst upper-bound than in the accepted answer, namely $d^{-1+o(1)}$ instead of $d^{-1}$. 

>The advantage of the new method is that it can be applied to (almost all) one-dimensional projections of arbitrary isotropic log-concave distributions (thanks to Klartag's CLT for convex bodies, namely [Theorem 1.3 of this paper][1]).

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Let $F$ be the marginal cdf of $\sqrt{d}x_1$ and let $F^c := 1 - F$ be its survival function. Let $\Phi$ be the standard gaudssian cdf.

We will make use of the following fact (see ref https://mathoverflow.net/a/315232/78539)

>**Fact 1 (Approximation).** *$\sup_{t \in \mathbb R}|F(t) - \Phi(t)| \le C/d$, for an absolute constant $C>0$ which doesn't depend on $d$*.

We shall also need the following fact (which follows from Gaussian concentration of $N(0,1)$ and sub-exponential concentration of $x_1$)

>**Fact 2 (Concentration).** There exists an absolute constant $b>0$ such that $\max(F(-t),\Phi(-t),F^c(t),\Phi^c(t)) \le e^{-bt}$ for sufficiently large $t>0$.

Now, for any $T > 0$, one computes
$$
W_1(\sqrt{d}x_1,N(0,1)) = \int_{-\infty}^\infty |F(t)-\Phi(t)|dt = A_1(T) + A_2(T) + A_3(T),
$$
where the first is a a classical result (e.g see **Proposition 2.17** of [Santambrogio's OTAM][2]), and the $A_k(T)$'s are defined by
$$
\begin{split}
A_1(T) &:= \int_{-\infty}^{-T}|F(t)-\Phi(t)|dt,\\
A_2(T) &:= \int_{-T}^T|F(t)-\Phi(t)|dt,\\
A_3(T) &:= \int_{T}^\infty|F(t)-\Phi(t)|dt=\int_{T}^\infty|F^c(t)-\Phi^c(t)|dt.
\end{split}
$$

Thanks to **Fact 1**, we know that $A_2(T) = \mathcal O(T/d)$. On the other hand, thanks to **Fact 2**, we have for sufficiently large $T>0$,
$$
A_1(T) \le \int_{-\infty}^{-T} \max(F(t),\Phi(t))dt \le \int_{-\infty}^{-T} e^{bt}dt = \frac{e^{-bT}}{b}.
$$

By a symmetric argument, we also have $A_3(T) \le e^{-bT}/b$.
Taking $T=(\log d)/b$ then gives $A_2(T) = \mathcal O((\log d)/d)$ and $A_1(T),A_3(T) = \mathcal O(1/d)$.
Thus 
$$
W_1(x_1,N(0,1/d)) = W_1(\sqrt{d} x_1,N(0,1)) = \mathcal O(\frac{\log d}{d})=\mathcal O(d^{-1+o(1)}).
$$


  [1]: https://arxiv.org/pdf/math/0605014.pdf
  [2]: http://math.univ-lyon1.fr/~santambrogio/OTAM-cvgmt.pdf