No.
In the positive direction, Bourbaki proves in Topologie Générale, Chapitre IX, §9, Proposition 4 the following:
Let $G$ be a metrizable topological group. If $H$ is a closed normal subgroup then $G/H$ is metrizable. If $G$ is complete (in one of the one-sided uniformities) then so is $G/H$.
This entails a positive answer for metrizable abelian groups, as the left, right and two-sided uniformities coincide.
On the other hand, in Espaces Vectoriels Topologiques, Chapitre IV, §4, Exercices 9 et 10, they give among many other things a (somewhat involved) construction of a complete topological vector space whose quotient by a closed subspace is not complete. The linear structure is of no importance for the topological/uniform considerations, so this applies to topological abelian groups as well.
Worse, still: Susanne Dierolf proved in Über Quotienten vollständiger topologischer Vektorräume, Manuscripta Mathematica 17, Nr 1 (1975), 73–77 that every topological vector space arises as a quotient of a complete and Hausdorff topological vector space.