What is the universal property of associated graded?
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 ⊕Ni to ⊕Ni/Ni-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.
The associated graded of a filtered R-module M is the universal R-module with a map of the Rees module of M over R[t] to gr M.
Let me explain what the Rees module Rees(M) is: it's the submodule of M[t,t-1] which is generated as a R[t] module by tiM_i. Give this the obvious grading by degree of t. So Rees(M)/tRees(M)=gr M, whereas Rees(M)/(t-1)Rees(M)=M with the induced filtration. This is the thing that has a map to gr M.
The associated graded functor has an obvious universal property if you use a sufficiently nice definition of the notion of "being filtered". A good notion of the category of filtered objects over a category $\mathcal{C}$ consists of the functor category $\text{Fun}((\mathbb{Z},\leq), \mathcal{C})$, where the poset $(\mathbb{Z},\leq)$ is viewed as a category. In other words you just give yourself the filtration pieces $V_i$ and arbitrary maps $V_i \rightarrow V_{i+1}$ instead of just monomorphisms.
This category has a tensor product via Day convolution, and the dualizable objects (if $\mathcal{C}$ is say the category of vector spaces over a field) essentially correspond to the classical filtered vector spaces, via $V = \text{colim}_i V_i$
The associated graded functor then simply is the left adjoint to the "trivial filtration functor", sending a graded vector space $(V_i)_{i \in \mathbb{Z}}$ to the filtered vector space with $V_i \rightarrow V_{i+1}$ being the zero map.