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)^2 &=2 u - (p-1) p u^{p-1} + 2 (p-2) p u^{p} - (2 - 3 p + p^2) t^{1 + p} \\ &>2 u - (p-1) p u^{p-1} + 2 (p-2) p u^{p} - (2 - 3 p + p^2) u^{1 + p} \\ &=:f(u). \end{align*} if $0<t<u$. We have \begin{equation} f''(u)=p \left(p^2-3 p+2\right) (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.