Lemma 1.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.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*}
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. $\qed$
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. $\qed$
Remark 1. The lower bound on $\|x\|_p$ or, equivalently, on $m_p$, is an easy corollary of H"older'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 $n\to\infty$. $\qed$
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 $i=1,\dots,n$.
