A universal property comes from an adjunction. From this point of view, associated graded has no universal property because it is not left or right adjoint.

Proof. If gr(-) were left (right)
adjoint, then it would respect
cokernels (kernels). Consider the
morphism of filtered vector spaces
(0⊆0⊆V)→(0⊆V⊆V)
(the three pieces are the 0-, 1-, and
2-filtered parts) which is just the
identity map on V. It's kernel and
cokernel are trivial. But the induced
map
gr(0⊆0⊆V)→gr(0⊆V⊆V)
is the zero map from V (in degree 2)
to V (in degree 1), which has
non-trivial kernel and cokernel. So
the associated graded of the
(co)kernel is not the (co)kernel of
the associated graded map.

Ben's solution is to write this poorly behaved functor as a composition of two nicer functors. The first functor is Rees:R-filmod→R[t]-grmod (from the category of filtered R-modules to the category of graded R[t]-modules). I think this functor is right adjoint to R[t]/(t-1)⊗-.

The second is R[t]/(t)⊗-:R[t]-grmod→R-grmod, the functor that takes ⊕N_{i} to ⊕N_{i}/N_{i-1}. R[t]/(t)⊗- is left adjoint to the functor that takes a graded R-module to the same graded module, regarded as an R[t]-module by letting t act by 0.

**Upshot:** associated graded is not an adjoint functor, so it doesn't have a nice universal property by itself, but it is the composition of a right adjoint functor and a left adjoint functor, which do have universal properties.