Does the usual filtration on graded objects satisfy a universal property? Let $\mathsf{A}$ be an abelian category, such as $R \mathsf{Mod}$ for concreteness. We can think of the category of $\mathbb{Z}$-graded objects in $\mathsf{A}$ as the functor category $\mathsf{A}^\mathbb{Z}$ where $\mathbb{Z}$ is viewed as a discrete category. This works for $\mathbb{N}$-gradings as well. 
The category of filtered objects in $\mathsf{A}$ in the classical sense may (potentially) be too small, so we can allow "generalized" filtrations as Nicolas Schmidt suggested in his answer to a related question. Then the generalized filtrations are the objects of the functor category $\mathsf{A}^{(\mathbb{Z}, \leq)}$, where $(\mathbb{Z}, \leq)$ is the ordered group of integers viewed as a poset category. We could of course use the poset $(\mathbb{N}, \leq)$ instead if we need to.
Now, given an a graded object $A$ we can form the corresponding filtered object with pieces $A^{(n)} = \bigoplus_{i \leq n} A_i$. This is obviously a functor from graded objects in $\mathsf{A}$ to filtered objects in $\mathsf{A}$. Indeed, given a morphism $f: A \to B$ of graded objects, which we write in degree $i$ as $f_i: A_i \to B_i$, the corresponding morphism of filtered objects is the obvious one $f^{(n)} := \bigoplus_{i \leq n} f_i : A^{(n)} \to B^{(n)}$.
Now, does this "filtration" functor satisfy a universal property? That is, does it have a left or right adjoint? 
 A: Let us call your functor $F$.  Then $F$ is left adjoint to the functor $G:\mathsf{A}^{(\mathbb{Z},\leq)}\to \mathsf{A}^\mathbb{Z}$ which is just composition with the inclusion functor $i:\mathbb{Z}\to(\mathbb{Z},\leq)$.  That is, $G$ takes a filtered object to the graded object whose graded pieces are the objects in the filtration.
This is straightforward to check directly, but you can also see it by some general nonsense.  Indeed, the left adjoint to $G$ is by definition the functor that takes a functor $\mathbb{Z}\to \mathsf{A}$ to its left Kan extension along the inclusion $i:\mathbb{Z}\to(\mathbb{Z},\leq)$.  Your formula for $F$ is then just the usual colimit formula for computing left Kan extensions: to compute the left Kan extension of $T:\mathbb{Z}\to\mathsf{A}$ at an object $n\in(\mathbb{Z},\leq)$, you form the colimit of the diagram formed by objects $T(m)$ indexed by the comma category $(i\downarrow n)$.  That comma category is just the discrete category of all $m\leq n$, so this colimit is your direct sum $\bigoplus_{m\leq n} T(m)$.
(Of course, there is nothing special about $\mathbb{Z}$ here--the same thing would work for any poset.)
