By rescaling, without loss of generality (wlog) $\|x\|_\infty=1$. 
Let $X$ be a random variable such that $P(X=|x_i|)=1/n$ for $x=(x_1,\dots,x_n)$ and all $i=1,\dots,n$. Then 
\begin{equation*}
	\|x\|_r^r=nm_r,\quad m_r:=EX^r, 
\end{equation*}
for $r\ge1$. Let us find the best possible bounds on $m_p$ in terms of $m_0=1,m_1,m_2$; these bounds can then be easily translated back in the terms of $\|x\|_r$. Wlog 
\begin{equation*}
0<m_1<\sqrt{m_2}<\sqrt{m_1}<1. \tag{1}	
\end{equation*}
Everywhere here $0\le t,u\le1$. 

>Lemma 1. $d_1(t):=(p-1)  u^{p-2}t^2+(2-p) u^{p-1}t-t^p\le0$, and $d_1(t)=0$ if $t\in\{0,u\}$. 

>Lemma 2. $d_2(t):=c t^2+b t+a-t^p\ge0$, and $d_2(t)=0$ if $t\in\{u,1\}$, where 
\begin{align*}
	a&:=\frac{(p-2) u^{p+1}-(p-1) u^p+u^2}{(1-u)^2}, \\
	b&:=\frac{p u^{p-1}-(p-2) u^{p+1}-2 u}{(1-u)^2}, \\ 
	c&:=\frac{-p u^{p-1}+(p-1) u^p+1}{(1-u)^2}. 
\end{align*}

These lemmas will be proved at the end of this answer. 

It follows from Lemma 1 that 
\begin{equation*}
	0\ge Ed_1(X)=(p-1)  u^{p-2}m_2+(2-p) u^{p-1}m_1-m_p, \tag{2}
\end{equation*}
and this inequality turns into the equality if $u=u_*:=m_2/m_1$ and $P(X=u_*)=m_1^2/m_2=1-P(X=0)$; in view of (1), such a r.v. $X$ exists, and $u_*\in(0,1)$. So, substituting $u=u_*$ into (2), we have the best possible lower bound on $m_p$: 
\begin{equation}
	m_p\ge m_1(m_2/m_1)^{p-1}. 
\end{equation}

Similarly, it 
It follows from Lemma 2 that 
\begin{equation*}
	0\le Ed_2(X)=c m_2^2+b m_1+a-m_p, \tag{3}
\end{equation*}
and this inequality turns into the equality if $u=u_{**}:=(m_1-m_2)/(1-m_1)$ and $P(X=u_{**})=(1-m_1)/(1-u_{**})=1-P(X=1)$; in view of (1), such a r.v. $X$ exists, and $u_{**}\in(0,1)$. So, substituting $u=u_{**}$ into (3), we have the best possible upper bound on $m_p$: 
\begin{equation}
	m_p\le \frac{m_2-m_1^2+(1-m_1)^2 (\frac{m_1-m_2}{1-m_1})^p}{1+m_2-2 m_1}. 
\end{equation}

------------------

*Proof of Lemma 1.* We have $d_1(0)=0=d_1(u)=d'_1(u)$. Also, $d''_1(t)=(p-1) (2 u^{p-2} - p t^{p-2})$ switches in sign from $+$ to $-$ (only once) as $t$ increases from $0$ to $1$, and the switch point is $<u$. Now Lemma 1 follows. 

*Proof of Lemma 2.* We have 
\begin{align*}
	d''_2(t)(1-u)^2u &=2 \left((p-1) u^{p+1}-p u^p+u\right)-(p-1) p (1-u)^2 u t^{p-2} \\ 
&>2 \left((p-1) u^{p+1}-p u^p+u\right)-(p-1) p (1-u)^2 u^{p-1} \\
 &=:f(u)  
\end{align*}
if $0<t<u$. 
We have 
\begin{equation}
	f''(u)=(p-2) (p-1) p (1-u) u^{p-3} ((p+1)u-(p-1)), 
\end{equation}
so that $f''(u)$ switches in sign from $-$ to $+$ (only once) as $u$ increases from $0$ to $1$. Also, $f(0)=f(1)=f'(1)=0$. So, $f>0$ and hence $d''_2(t)>0$ if $0<t<u$. Also, clearly $d_2$ has at most one inflection point. Also, $d_2(1)=0=d_2(u)=d'_2(u)$. Now Lemma 2 follows.