If $L$ is a solvable Lie algebra over an algebraically closed field of characteristic zero, then an abelian subalgebra of maximal dimension in $L$ has the same dimension as an abelian *ideal* of maximal dimension in $L$, see Proposition $2.6$ [here](http://homepage.univie.ac.at/Dietrich.Burde/papers/burde_39_max_ab.pdf). For an ideal, the question is easier to decide than for a subalgebra. However in general, 
to decide whether or not there is an abelian subalgebra of given dimension in $L$
the algorithms mentioned use Gröbner bases in one or the other way. <br>
On the other hand, for semisimple Lie algebras the maximal dimension of an abelian subalgebra is known, see the article "Abelian ideals in a Borel subalgebra of a complex simple Lie algebra" by R. Suter (2004) and the paper cited above.