$\newcommand{\R}{\mathbb R}$This is only a partial answer in the sense that it will provide $\frac 1p$-Hölder maps, not Lipschitz. I still hope it can help.
Fix a big radius $R>0$ (I like $R\to\infty$ instead of $n$, it helps me to better visualize), and consider the map $T=T_R:\R^d\to\R^d$ projecting points onto the ball $B_R=B_R(0)\subset\R^d$ given by
$$
T_R(x)=\operatorname{Proj}_{B_R}(x)=
\begin{cases}
x & \text{if }|x|\leq R\\
R\frac{x}{|x|} & \text{if }|x|\geq R.
\end{cases}
$$
Consider $I_R:W_1(\R^d)\to W_p(\R^d)$ defined as
$$
I_R(\mu):= {T_R}_{\#}\mu.
$$
Since $T_R$ maps $\R^d$ to $\overline{B_R}$ clearly the measure $I(\mu)$ has support in $\overline{B_R}$ and thus has finite $p$-moment for any $p>1$, i-e $I_R$ indeed maps $W_1$ to $W_p$.
I first claim that $I_R$ is $\frac 1p$-Hölder. For this, fix $\mu,\nu\in W_1$ and let $\pi\in\Pi(\mu,\nu)$ be any optimal plan for the $W_1$ transportation problem betwenn $\mu,\nu$. For convenience let me write $\mu_R=I_R(\mu)$ and $\nu_R=I_R(\nu)$, and set $\pi_R:=(T_R,T_R)_{\#}\pi$. It is immediate to check that $\pi_R\in \Pi(\mu_R,\nu_R)$ has marginals $\mu_R,\nu_R$, hence
$$
W_p^p(\mu_R,\nu_R)\leq \iint |x-y|^p\pi_R(dx,dy)=\iint |T_R(x)-T_R(y)|^p\pi(dx,dy).
$$
It is easy to check that $|T_R(x)-T_R(y)|\leq |x-y|$ ($T_R$ is the projection onto the convex set $B_R$!), hence writing
\begin{multline*}
|T_R(x)-T_R(y)|^p=|T_R(x)-T_R(y)|^{p-1}|T_R(x)-T_R(y)|
\\
\leq (2R)^{p-1}|T_R(x)-T_R(y)|\leq (2R)^{p-1}|x-y|
\end{multline*}
we get immediately
$$
W_p^p(\mu_R,\nu_R)\leq (2R)^{p-1}\iint |x-y|\pi(dx,dy)=(2R)^{p-1}W_1(\mu,\nu).
$$
(This is a classical trick allowing to interpolate $W_p$ between $W_1$ and $W_\infty$, roughly speaking.)
This reads
$$
W_p(I(\mu),I(\nu))\leq (2R)^{\frac{p-1}{p}} W_1^{\frac 1p}(\mu,\nu)
$$
and thus $I=I_R$ is $\frac 1p$-Hölder (with quantified Hölder constant).
Let me finally check that, for any fixed $\mu\in W_1$ we have convergence $I_R(\mu)=\mu_R\to \mu$ as $R\to+\infty$.
For this, note that $\pi_R:=(\operatorname{id},T_R)_{\#}\mu$ has marginals $\mu, \mu_R$. As a consequence
\begin{multline*}
W_1(\mu,\mu_R)
\leq
\iint |x-y|\pi_R(dx,dy)
=\int_{\R^d} |x-T_R(x)|\mu(dx)
=\int_{|x|\geq R}\left|x-R\frac{x}{|x|}\right|\mu(dx)
\\
=\int_{|x|\geq R}\left|1-\frac{R}{|x|}\right||x|\mu(dx)
\leq
2\int_{|x|\geq R}|x|\mu(dx).
\end{multline*}
(The las inequality follows from $|1-\frac{R}{|x|}|\leq 1 + \frac{R}{|x|}\leq 2$ for $|x|\geq R$.)
Finally, since $\mu\in W_1$ has finite first moment $\int|x|\mu(dx)<\infty$ it follows that the last integral converges to zero as $R\to\infty$ and the claim follows.
Comment: resorting to compact support in $B_R$ is very rough and ends up giving $\frac 1p$ Hölder, bu you may actually get Lipschitz by refining the idea, for example considering the map
$$
T_R(x)=\begin{cases}
x & \text{if }|x|\leq R\\
R\left|\frac{x}{R}\right|^{\frac 1p}\frac{x}{|x|} & \text{if }|x|\geq R.
\end{cases}
$$
transforms 1st moments of $\mu$ into p-th moments of $\mu_R={T_R}_{\#}\mu$. I didn't check the rest of the details, but if you really want Lipschitz it's worth a try.
(Certainly some convexity should help: For example $T_R$ is the gradient of a convex map, so it is automatically an optimal map from $\mu$ to $\mu_R$ and the last step to prove that $W_1(\mu,\mu_R)\to 0$ works exactly in the same way.)
