$\newcommand{\de}{\delta}$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. $\quad\Box$