$\newcommand\T{\mathrm T}$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 $\T V \to S$.

In the original version of this answer, I proceeded as follows.  For each $v^* \in V^* \mathrel{:=} \operatorname{Hom}_\text{\(R\)-mod}(V^*, R)$, let $\alpha_{v^*}$ be the unique extension of the map $V \to R[x]$ given by $v \mapsto \langle v^*, v\rangle x$ to an $R$-module map $\T V \to R[x]$.  Then, for each $n \in \mathbb Z_{\ge 0}$, $\T^nV$ is the set of $\tau \in \T V$ such that $\alpha_{v^*}(\tau)$ lies in $R x^n$ for all $v^* \in V^*$.

This definition is too coarse: it correctly identifies the degree of each homogeneous element, but, as you [pointed out](https://mathoverflow.net/questions/405276/proving-the-graded-structure-of-the-tensor-algebra-from-only-the-universal-prope/442828#comment1142591_405279), it can also incorrectly assign elements degrees that they shouldn't have.

The obvious fix is to replace $R$-module maps $V \to R$ by $R$-module maps $V \to A$ for arbitrary $R$-algebras $A$.  This works, but the only way that I can see to show that it always works is to take $A$ to be the usual explicitly constructed tensor algebra.