norm inequalities
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+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}
It remains to prove Lemmas 1 and 2.
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. $\Box$
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. $\Box$
Remark 1. The lower bound on $\|x\|_p$ or, equivalently, on $m_p$, is an easy corollary of Hölder's inequality or, equivalently, of the log-convexity of $\|x\|_r$ in $r$. However, in this case, it seemed to make sense to use a similar approach to both the lower bound and (the much more difficult) upper bound.
Remark 2. The bounds in terms of $m_r$ seem more natural than those in terms of $\|x\|_r$ (even though these two kinds of bounds are very easy to translate into each other). One reason for this preference is that restrictions (1) when translated into the $\|x\|_r$ terms will contain $n$. Moreover, the restrictions $m_1<\sqrt{m_2}$ and $\sqrt{m_1}<1$ in (1), when translated into the $\|x\|_r$ terms, become rather restrictions on $n$: $n>\|x\|_1^2/\|x\|_2^2$ and $n>\|x\|_1$, respectively. In these remarks, it is still assumed that $\|x\|_\infty=1$.
Remark 3. The lower bound on $m_p$, when translated into the $\|x\|_r$ terms, is free of $n$: \begin{equation} \|x\|_p^p\ge\|x\|_2^{2p-2}/\|x\|_1^{p-2}. \end{equation}
However, if one insists on a free of $n$ upper bound on $\|x\|_p$ in terms of $p$, $\|x\|_1$, and $\|x\|_2$ only, then the bound cannot be substantially better than the trivial upper bound given by the inequality $\|x\|_p^p\le\|x\|_2^2$, as in the first display in the OP. Indeed, we have
Proposition 1. Suppose that \begin{equation} \|x\|_p\le F(\|x\|_1,\|x\|_2^2) \end{equation} for all vectors $x$, where the function $F=F_p$ is continuous (and does not depend on $n$). Then \begin{equation} F(A,B)\ge B^{1/p} \end{equation} for all natural $B$ and all real $A>B$.
The restriction $A>B$ corresponds to the restriction $\sqrt{m_2}<\sqrt{m_1}$ in (1).
Proof of Proposition 1. Take indeed any natural $B$ and any real $A>B$. For natural $n>A$, let $y=(y_1,\dots,y_n)$, where $y_1=\dots=y_B=1$ and $y_{B+1}=\dots=y_n=\frac{A-B}{n-B}[\in(0,1)]$. Then $\|y\|_1=A$, $\|y\|_2^2\to B$, $\|y\|_p^p\to B$, and hence \begin{equation} B^{1/p}\leftarrow\|y\|_p\le F(\|y\|_1,\|y\|_2^2)\to F(A,B) \end{equation} as $n\to\infty$. $\Box$
In particular, it follows from Proposition 1 that the conjecture $$ \|x\|_p\le C_p\|x\|_2\Big(\frac{\|x\|_\infty}{\|x\|_1}\Big)^r$$ with $r=(p-2)/(2p-2)$, proposed by the OP, is false in general. This can also be seen directly by letting $x_i=1$ for all $i=1,\dots,n$.
A likely candidate for the optimal solution is when the $x_i$ take $3$ different values, say $x_i = a$ for $i=1 \ldots n_1$, $b$ for $i=n_1+1 \ldots n_1 + n_2$, $\|x\|_\infty$ for $i=n_1+n_2+1 \ldots n$, where $n_i \ge 0$, $ n_1 + n_2 \le n$, and $a$ and $b$ must satisfy
$$ \eqalign{ n_1 a + n_2 b + (n-n_1-n_2) \|x\|_\infty &= \|x\|_1\cr
n_1 a^2 + n_2 b^2 + (n-n_1-n_2) \|x\|_\infty^2 &= \|x\|_2^2\cr
0 \le a,b &\le \|x\|_\infty
} $$
For given $n, n_1, n_2, \|x\|_1, \|x\|_2, \|x\|_\infty$, this will determine $a$ and $b$ (up to interchange) as the roots of a quadratic, when that quadratic has
two roots in the interval $[0, \|x\|_\infty]$. Which $n_1$ and $n_2$ give the maximum or minimum value of $\|x\|_p$ will likely vary.