$\newcommand{\de}{\delta}$The answer to your question is negative if $p<2$.

Indeed, let $\nu_i=\de_0$ for all $i$, where $\de_a$ is the Dirac measure supported on the singleton set $\{a\}$. Let $X_1,\dots,X_n$ be independent random variables (r.v.'s) with respective distributions $\mu_1,\dots,\mu_n$. Let $Y_i:=X_i^2$ for all $i$. Let $\mu:=\bigotimes_1^n\mu_i$ and $\nu:=\bigotimes_1^n\nu_i$.

Then $W_p(\mu_i,\nu_i)=(E|X_i|^p)^{1/p}$ and $W_p(\mu,\nu)=(E(\sum_1^n X_i^2)^{p/2})^{1/p}$.

So, the inequality in question becomes \begin{equation} L:=\Big\|\sum_1^n Y_i\Big\|_{p/2}\overset{\text{(?)}}\le \sum_1^n \|Y_i\|_{p/2}=:R. \end{equation} Suppose now that $n=2$ and $Y_1,Y_2$ are independent r.v.'s such that $P(Y_i=1)=t=1-P(Y_i=0)$ for $i=1,2$, where $t\downarrow0$. Then $L=(2t(1-t)+t^2 2^{p/2})^{2/p}\sim(2t)^{2/p}$, whereas $R=2t^{2/p}$, so that the inequality $L\le R$ fails to hold if $p<2$ and $t$ is small enough. $\quad\Box$

$\newcommand{\de}{\delta}\renewcommand{\S}{\mathcal S}\newcommand{\T}{\mathcal T}$The answer to your question is negative if $p<2$.

Indeed, let $\nu_i=\de_0$ for all $i$, where $\de_a$ is the Dirac measure supported on the singleton set $\{a\}$. Let $X_1,\dots,X_n$ be independent random variables (r.v.'s) with respective distributions $\mu_1,\dots,\mu_n$. Let $Y_i:=X_i^2$ for all $i$. Let \begin{equation*} \text{$\mu:=\bigotimes_1^n\mu_i$ and $\nu:=\bigotimes_1^n\nu_i$.} \end{equation*}

Then $W_p(\mu_i,\nu_i)=(E|X_i|^p)^{1/p}$ and $W_p(\mu,\nu)=(E(\sum_1^n X_i^2)^{p/2})^{1/p}$.

So, the inequality in question becomes \begin{equation*} L:=\Big\|\sum_1^n Y_i\Big\|_{p/2}\overset{\text{(?)}}\le \sum_1^n \|Y_i\|_{p/2}=:R. \end{equation*} Suppose now that $n=2$ and $Y_1,Y_2$ are independent r.v.'s such that $P(Y_i=1)=t=1-P(Y_i=0)$ for $i=1,2$, where $t\downarrow0$. Then $L=(2t(1-t)+t^2 2^{p/2})^{2/p}\sim(2t)^{2/p}$, whereas $R=2t^{2/p}$, so that the inequality $L\le R$ fails to hold if $p<2$ and $t$ is small enough. $\quad\Box$

I have just gotten up during the night and it occurred to me how to show, very simply, that the inequality in question holds for $p\ge2$, and then I saw the answer by the OP :-), which is correct, except it had to be mentioned there that the pairs $(X_i,Y_i)$ may be assumed to be independent.

Actually, we have the following somewhat more general fact. Let $p\in[2,\infty)$. For each $i\in[n]:=\{1,\dots,n\}$, let $(S_i,\S_i,\mu_i)$ and $(T_i,\T_i,\nu_i)$ be probability spaces. Let $(S,\S,\mu):=\bigotimes_{i\in[n]}(S_i,\S_i,\mu_i)$ and $(T,\T,\nu):=\bigotimes_{i\in[n]}(T_i,\T_i,\nu_i)$. For each $i\in[n]$, let $\Pi(\mu_i,\nu_i)$ denote the set of all probability measures on the measurable space $(S_i\times T_i,\S_i\otimes\T_i)$ with marginals $\mu_i$ and $\nu_i$. Let $\Pi(\mu,\nu)$ denote the set of all probability measures on the measurable space $(S\times T,\S\otimes\T)$ with marginals $\mu$ and $\nu$. For each $i\in[n]$, let $f_i\colon S_i\times T_i\to[0,\infty)$ be a measurable function. Let $f(s,t):=\sqrt{\sum_{i\in[n]} f_i(s_i,t_i)^2}$ for all $s=(s_1,\dots,s_n)\in S= S_1\times\cdots\times S_n$ and $t=(t_1,\dots,t_n)\in T=T_1\times\cdots\times T_n$. For each $i\in[n]$, let \begin{equation*} W_p(\mu_i,\nu_i):=\inf_{\pi_i\in\Pi(\mu_i,\nu_i)} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{1/p}. \end{equation*} Let \begin{equation*} W_p(\mu,\nu):=\inf_{\pi\in\Pi(\mu,\nu)} \Big(\int_{S\times T}f(s,t)^p\,\pi(ds,dt)\Big)^{1/p}. \end{equation*} Then \begin{equation*} W_p(\mu,\nu)\overset{\text{(?)}}\le \Big(\sum_{i\in[n]} W_p(\mu_i,\nu_i)^2\Big)^{1/2}. \tag{1}\label{1} \end{equation*}

Indeed, take any real $w_1,\dots,w_n$ such that $W_p(\mu_i,\nu_i)<w_i$ (if such $w_1,\dots,w_n$ exist; otherwise, inequality \eqref{1} is trivial). Then for each $i\in[n]$ there exists some $\pi_i\in\Pi(\mu_i,\nu_i)$ such that \begin{equation*} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{1/p}<w_i. \tag{2}\label{2} \end{equation*} Let $$\pi:=\bigotimes_{i\in[n]}\pi_i.$$ Then $\pi\in\Pi(\mu,\nu)$. Moreover, for each $i\in[n]$ \begin{equation*} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{2/p} =\|g_i\|_{L^{p/2}(\pi)}, \end{equation*} where $g_i(s,t):=f_i(s_i,t_i)^2$ for for all $s=(s_1,\dots,s_n)\in S= S_1\times\cdots\times S_n$ and $t=(t_1,\dots,t_n)\in T=T_1\times\cdots\times T_n$, and \begin{equation*} \Big(\int_{S\times T}f(s,t)^p\,\pi(ds,dt)\Big)^{2/p} =\|g\|_{L^{p/2}(\pi)}, \end{equation*} where $g:=f^2=g_1+\cdots+g_n$.

So, by the condition $p\in[2,\infty)$, Minkowski's inequality, and \eqref{2}, \begin{multline*} W_p(\mu,\nu)^2\le \Big(\int_{S\times T}f(s,t)^p\,\pi(ds,dt)\Big)^{2/p} =\|g\|_{L^{p/2}(\pi)} \\ \le\sum_{i\in[n]} \|g_i\|_{L^{p/2}(\pi)} =\sum_{i\in[n]} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{2/p} <\sum_{i\in[n]}w_i^2. \end{multline*} Letting now $w_i\downarrow W_p(\mu_i,\nu_i)$ for each $i\in[n]$, we get \eqref{1}. $\quad\Box$

$\newcommand{\de}{\delta}$The answer to your question is negative if $p<2$.

Indeed, let $\nu_i=\de_0$ for all $i$, where $\de_a$ is the Dirac measure supported on the singleton set $\{a\}$. Let $X_1,\dots,X_n$ be independent random variables (r.v.'s) with respective distributions $\mu_1,\dots,\mu_n$. Let $Y_i:=X_i^2$ for all $i$.

Then the inequality in question becomes \begin{equation} L:=\Big\|\sum_1^n Y_i\Big\|_{p/2}\overset{\text{(?)}}\le \sum_1^n \|Y_i\|_{p/2}=:R. \end{equation} Suppose now that $n=2$ and $Y_1,Y_2$ are independent r.v.'s such that $P(Y_i=1)=t=1-P(Y_i=0)$ for $i=1,2$, where $t\downarrow0$. Then $L=(2t(1-t)+t^2 2^{p/2})^{2/p}\sim(2t)^{2/p}$, whereas $R=2t^{2/p}$, so that the inequality $L\le R$ fails to hold if $p<2$ and $t$ is small enough. $\quad\Box$

$\newcommand{\de}{\delta}$The answer to your question is negative if $p<2$.

Indeed, let $\nu_i=\de_0$ for all $i$, where $\de_a$ is the Dirac measure supported on the singleton set $\{a\}$. Let $X_1,\dots,X_n$ be independent random variables (r.v.'s) with respective distributions $\mu_1,\dots,\mu_n$. Let $Y_i:=X_i^2$ for all $i$. Let $\mu:=\bigotimes_1^n\mu_i$ and $\nu:=\bigotimes_1^n\nu_i$.

Then $W_p(\mu_i,\nu_i)=(E|X_i|^p)^{1/p}$ and $W_p(\mu,\nu)=(E(\sum_1^n X_i^2)^{p/2})^{1/p}$.

So, the inequality in question becomes \begin{equation} L:=\Big\|\sum_1^n Y_i\Big\|_{p/2}\overset{\text{(?)}}\le \sum_1^n \|Y_i\|_{p/2}=:R. \end{equation} Suppose now that $n=2$ and $Y_1,Y_2$ are independent r.v.'s such that $P(Y_i=1)=t=1-P(Y_i=0)$ for $i=1,2$, where $t\downarrow0$. Then $L=(2t(1-t)+t^2 2^{p/2})^{2/p}\sim(2t)^{2/p}$, whereas $R=2t^{2/p}$, so that the inequality $L\le R$ fails to hold if $p<2$ and $t$ is small enough. $\quad\Box$