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).** *$Ko(x_1,N(0,1)) := \sup_{t \in \mathbb R}|F(t) - \Phi(t)| \le C/d$, for an absolute constant $C>0$ which doesn't depend on $d$*.

We shall also need the following fact (which follows concentration of $\mathcal O(1)$-Lipschitz transformations of log-concave random vectors)

>**Fact 2 (Concentration).** There exists an absolute constant $b>0$ such that $\max(F(-t),\Phi(-t),F^c(t),\Phi^c(t)) \le e^{-bt^2}$ 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 $T>0$,
$$
A_1(T) \le \int_{-\infty}^{-T} \max(F(t),\Phi(t))dt \le \int_{-\infty}^{-T} e^{bt^2}dt = \mathcal O(e^{-bT^2}).
$$

By a symmetric argument, we also have $A_3(T) = \mathcal O(e^{-bT^2})$.
Taking $T=\sqrt{(\log d)/b}$ then gives $A_2(T) = \mathcal O(\sqrt{\log d}/d)$ and $A_1(T),A_3(T) = \mathcal O(1/d)$.
Thus 
$$
W_1(x_1,N(0,1/d)) = W_1(\sqrt{d} x_1,N(0,1)) = \mathcal O(\frac{\sqrt{\log d}}{d})=\mathcal O(d^{-1+o(1)}).
$$

Extension to one-dimensional projections of isotropic log-concave distributions
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Let $x$ be a random vector in $\mathbb R^d$ with an arbitrary isotropic log-concave density. Recall that by isotropy, we mean
- $\mathbb E [x] = 0$, and
- $\mathbb E |x^\top u|^2=1$ for any unit-vector $u \in \mathbb R^d$.

Now, for any unit-vector $u \in \mathbb R^d$, let $x_u := x^\top u$ be the projection of $x$ in the direction of $u$, and let $F_u$ be its cdf. Finally, let $\sigma_d$ be the uniform distribution on the unit-sphere in $\mathbb R^d$. We will prove the following

>**Theorem.** For a proportion $1-e^{-\mathcal O(d^{0.99})}$  of unit-vectors $u$ in $\mathbb R^d$ (w.r.t the uniform distribution on the unit-sphere in $\mathbb R^d$), it holds that
$$
W_1(x_u,N(0,1)) = \mathcal O((\log d)^{-1/2+o(1)})=o(1).
$$

Thanks to Theorem 1.3 of Klartag's paper (referenced in the preamble), we know that for $\varepsilon_d = \left(\dfrac{\log \log d}{\log d}\right)^{1/2}=o(1)$,

>**Fact 1' (Approximation).** There exists a measurable subset $\mathcal U$ of unit-vectors with $\sigma_d(\mathcal U) \ge 1 - e^{-\mathcal O(d^{0.99})}$ such that
$$
Ko(x_u,N(0,1)) \le \mbox{TV}(x_u,N(0,1)) = \mathcal O(\varepsilon_d)
$$
for all $u \in \mathcal U$.

Just as in **Fact 2**, we have the following property which is a direct consequence of the concentration of $\mathcal O(1)$-Lipschitz transformations of log-concave random vectors,

>**Fact 2' (Concentration).** There exists an absolute constant $b>0$ such that
$$
\max(F_u(-t),\Phi(-t),F_u^c(t),\Phi^c(t)) \le e^{-bt^2}
$$
for all $u \in \mathcal U$ and for sufficiently large $t>0$.

Emulating the proof of the spherical case, but using the cutoff $T=\sqrt{\log (1/\varepsilon_d)/b}$ instead, we obtain
$$
\sup_{u \in \mathcal U}W_1(x_u,N(0,1))) = \mathcal O(\varepsilon_d\sqrt{\log(1/\varepsilon_d)}) = \mathcal O((\log d)^{-1/2+o(1)}) = o(1).
$$
Noting that $\sigma_d(\mathcal U)\ge 1 - e^{-\mathcal O(d^{0.99})}$, the proof is complete. $\quad\quad\quad\quad\quad\quad\Box$

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