Edit. The undecidability result not exactly in my survey but can be deduced from it. Here is a correct reference: Perkins, Peter Unsolvable problems for equational theories. Notre Dame J. Formal Logic 8 1967 175–185. Perkins proves that there is no algorithm that, given a finite system of identities, says whether it is a basis of identities of a finite algebra (theorem 13). In fact he proves more. He constructs an algebra $E$ with undecidable word problem, and for every two terms $u,v$ of $E$ he constructs a finite set of identities $I(u,v)$ such that if $u=v$ in $E$ then the set $I(u,v)$ is the set of identities of a finite algebra, and if $u\ne v$, $I(u,v)$ holds on an infinite 1-generated algebra. Since we cannot decide whether $u=v$, we cannot decide, given a finite set of identities, it is a basis of a locally finite variety.
If you are interested in just one free object, the situation is even easier. It is known (Markov) that the finiteness of a 2-generated semigroup is undecidable. Now consider the signature consisting of the semigroup operation plus two 0-ary operations giving the generators. Then any finitely presented semigroup becomes a relatively free object in a variety given by a finite number of identities (involving the 0-ary operations). Thus it is undecidable, given a finite number of identities in that signature whether the 2-generated free algebra in the variety given by these identities is finite (that is almost exact quote from the survey).