Is the tensor product over $B$ of two flat $A$-modules flat over $A$? Given a morphism of commutative rings $A\to B$ such that $B$ is a flat $A$-module and given $M$, $N$ two $B$-modules flat as $A$-modules, is the tensor product $M\otimes_B N$ flat over $A$??
The tensor product $M\otimes_A N$ is flat over $A$, the proof is not hard:
Given an exact sequence $0\to X\to Y\to Z\to 0$ of $A$-modules since $M$ is
a flat $A$-module the sequence $0\to X\otimes_A M\to Y\otimes_A M\to Z\otimes_A M\to 0$ is exact. Again $N$ is a flat $A$-module, and since the tensor commute we have the statement.
I've tried some similar arguments but without any success, and I can't find a counterexample.
There is a morphism $M\otimes_A N\to M\otimes_B N$ (it is shown in here), but I don't know how to use it.

Assume that as $B$-modules, you have that $N= B\otimes_A N$: note that this is generally false. There is an obvious map $B\otimes_A N\to N$ that sends $b\otimes n\to bn$. One might be tempted to use the functional inverse $n\to 1\otimes n$, but this is not necessarily $B$-linear, unless say $A\to B$ is onto. If it were true, then the argument goes through. See this. As an example of how $B$ might definitely not be isomorphic to $B\otimes_A B$, take $B$ to be a $k$-algebra with $k$ a field, $k\to B$ the structure map, and say $B$ is finitely dimensional of dimension $>1$. Then $B\otimes_k B$ has dimension strictly greater than $B$. 
Now take $C$ an exact sequence. Then $N\otimes_A C$ is exact, and since $B$ is flat, so is $B\otimes_A N\otimes_A C$. Now write $$M\otimes_B N =(M\otimes_A B)\otimes_B (B\otimes_A N)$$
Then $M\otimes_B N\otimes C$ is obtained from $B\otimes_A N\otimes_A C$ by tensoring first over $B$ with $B$, which preserves exactness, and then with $M$ over $A$, which also does. 
