Unfortunately this will not work, but the counterexamples I know rely on results that are either difficult or unpublished.

1) Take $A_n = M_n{\mathbb C})$ with usual multiplication and $C^*$-norm. Each $A_n$ is amenable with constant $1$, so there exists an actual diagonal element in $A_n\hat\otimes A_n$ which has norm $1$. On the other hand, your algebra $B$ is an example of a non-nuclear $C^*$-algebra and hence it is non-amenable.

2) Take $A_n = \ell^1({\mathbb T}_d)$ where ${\mathbb T}_d$ denotes the circle group equipped with the discrete topology. Once again each $A_n$ is amenable with constant $1$. Many years ago I was told that B. E. Johnson and M. C. White had an unpublished result, observing that the corresponding algebra $B$ quotients onto the measure algebra $M({\mathbb T})$. Now it is is known by work of Brown and Moran that $M({\mathbb T})$ has a non-zero point derivation, and pulling this back to $B$ we see that $B$ is not weakly amenable, hence it is not approximately amenable.


Relatedly: knowing that a Banach algebra $A$ is amenable does not guarantee that $\ell^\infty({\mathbb N}; A)$ is amenable. For instance one can take $A={\mathcal K}(\ell_p)$, known to be amenable by work in Johnson's 1972 memoir; and then Runde has shown that $\ell^\infty({\mathbb N}; {\mathcal K}(\ell_p) )$ is non-amenable.