I realized that I didn't look at the definition careful enough. So I have erased my first answer, since it didn't say anything useful.

The requirements actually says that 

$(1)\quad \Delta(ab) = \Delta(a)b$

and 

$(2)\quad \Delta(ab) = a\Delta(b).$

In other words, the linear homomorphism $\Delta\colon A \to A\otimes_k A$ should be a homomorphism of $A$-$A$-bimodules.  Therefore the Frobenius dimension of an algebra is nothing else than the dimension of the space of homomorphisms

$\operatorname{Hom}_{A\otimes_k A^{\operatorname{op}}}(A,A\otimes_k A).$

This can be easily computed in QPA by the following line of commands:

    Aenv := EnvelopingAlgebra( A );
    P := DirectSumOfQPAModules( IndecProjectiveModules( Aenv ) );
    M := AlgebraAsModuleOverEnvelopingAlgebra( A );
    Length( HomOverAlgebra( M, P ) );

The last number is the Frobenius dimension of the algebra $A$.

So, Frobenius dimension = $\dim_k \operatorname{Hom}_{A\otimes_k A^{\operatorname{op}}}( A, A\otimes_k A )$.

The QPA-team.