My understanding is that the "right" way to define the symmetric algebra comes from a braiding that tells you how the symmetric group acts on tensor products.  And an easy and general way to get such a braiding is to consider the category of representations of a <a href="http://mathoverflow.net/questions/4640/are-supervector-spaces-the-representations-of-a-hopf-algebra">quasi-triangular Hopf algebra</a>; this is a short way to define the category of supervector spaces and recover the usual notion of graded-commutativity there.

The problem is that when you say "graded" (say, with respect to a group) you're only specifying at best a Hopf algebra, not a quasi-triangular structure.  So the answer depends on what possible quasi-triangular structures are floating around (in the $\mathbb{Z}/2\mathbb{Z}$ case on the group algebra of $\mathbb{Z}/2\mathbb{Z}$) and of course for a group of order $|G|$ terrible things are going to happen in characteristic dividing $|G|$.