I assume that "the universal property" means that, for every $R$-algebra $S$, every $R$-module map $V \to S$ extends uniquely to an $R$-algebra map $\operatorname TV \to S$.

Define $V \to \operatorname{Hom}_{\text{\(R\)-mod}}(V^*, R[x])$ by $v \mapsto v^* \mapsto \langle v^*, v\rangle x$.  Then, for each $n \in \mathbb Z_{\ge 0}$, $\operatorname T^n V$ is the pre-image under $\operatorname T V \to \operatorname{Hom}_{\text{\(R\)-mod}}(V^*, R[x])$ of $\operatorname{Hom}_{\text{\(R\)-mod}}(V^*, R x^n)$.

This is enough to define the grading.  It follows immediately that $\operatorname T V = \bigoplus_{n = 0}^\infty \operatorname T^n V$ when $V$ is finite-dimensional.  In general, we need to observe that the image of $V \to \operatorname{Hom}_{\text{\(R\)-mod}}(V^*, R[x])$ lies in the space of finite-rank homomorphisms.