Actually, there is a much stronger result, known as the Riesz-Thorin Theorem:
The subordinate norm $\|A\|_p$ is a log-convex function of $\frac1p$.
In other words,
$$\left(\frac1r=\frac\theta{p}+\frac{1-\theta}q\right)\Longrightarrow(\|A\|_r\le\|A\|_p^\theta\|A\|_q^{1-\theta}).$$
This contains as a particular case the inequality $\|A\|_2^2\le\|A\|_p\|A\|_{p'}$ that you mention. For a proof of R.-T. Theorem, see Section 7.3 (in the second edition) of my book Matrices, GTM216, Springer-Verlag (2010).
As for the last inequality, for $\|AB\|_2$, it is deadly false. If it was correct, then taking $B=A^T$, one would have
$$\|A\|_2^2=\|AA^T\|_2\le\|A\|_p\|A^T\|_{p'}=\|A\|_p^2,$$
hence $\|A\|_2\le\|A\|_p$ for every $p$ and $A$, which is obviously false. Indeed, take a rank-one matrix $A=xy^T$ ; then
$$\|A\|_p=\|x\|_p\|y\|_{p'}.$$
Take two vectors $x$ and $y$ such that $\|x\|_2=\|x\|_1$ and $\|y\|_2>\|y\|_\infty$ (possible if $n\ge2$), then $\|A\|_2 = \|x\|_2\|y\|_2> \|x\|_1\|y\|_\infty = \|A\|_1$.
Edit. Here is the elementary proof of $\|A\|_p\le\|A\|_1^{1/p}\|A\|_\infty^{1/p'}$. Recall the formulae
$$\|A\|_1=\max_j\sum_i|a_{ij}|,\qquad\|A\|_\infty=\max_i\sum_j|a_{ij}|.$$
Applying Hölder, one has
\begin{align}
\|Ax\|_p^p & = \sum_i\left|\sum_ja_{ij}x_j\right|^p\\&\le\sum_i\left(\sum_j|a_{ij}|\right)^{p/p'}\sum_k|a_{ik}|\,|x_k|^p \\
& \le \|A\|_\infty^{p-1}\sum_{i,k}|a_{ik}|\,|x_k|^p\\
&=\|A\|_\infty^{p-1}\sum_k|x_k|^p\sum_i|a_{ik}|\\
&\le\|A\|_\infty^{p-1}\|A\|_1\|x\|_p^p.
\end{align}
