In brief: such Gauss sums are easy to compute (when provided in a "fair" manner), and should be considered as invariants of the bicharacter (rather than the other way around).
In more detail: Building on work of Basak and Johnson, my collaborators and I showed in Appendix C here that there is a polynomial time algorithm to compute such a (homogeneous) quadratic Gauss sum. As you probably know, Wall classified irreducible metric groups. Basak and Johnson essentially pointed the way to an algorithmic classification of general metric groups into irreducible pieces. We did the complexity analysis of this approach. The Gauss sum of a reducible metric group is the product of the Gauss sums of its summands, and these can each be looked up in a table (due to Gauss, Wall,...) so that finishes the job.
For an alternative (and more general) approach, I quite enjoyed this paper by Cai et al.: On Tractable Exponential Sums.
It is perhaps worth stressing that when I say "polynomial time," I mean that A is a finite abelian group given explicitly in terms of its list of primary factors whose orders are written in binary. (On one hand, it would be unfair to require us to factorize A first; on the other hand, it would also be unfair if we could just sum over all elements of A directly.) We are given q as a matrix with integer entries.
As far as your last question (can we compute the sum from invariants of the metric group): the more helpful way for us to think about things is the converse. That is, the Gauss sum is itself an invariant of quadratic forms that packages up into a homomorphism from the Witt group to $U(1)$.