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$\newcommand\T{\mathrm T}$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, 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.

$\newcommand\T{\mathrm T}$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, 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, or something close to it. For example, I have already cheated a bit by using $R[x]$, which is just the tensor algebra on $R^*$; but I could cheat even more by taking for $A$ the quotient of the polynomial $R$-algebra in $V$-many indeterminates by $r x_v = x_{r v}$. Then there is a natural homomorphism $V \to A$ given by $v \mapsto x_v$, and we could use the above trick on $\T V \to A[x]$, or just observe that $A$ itself is graded (since we're quotienting out by a homogeneous ideal) and use the degree there. But at that point we've practically constructed the tensor algebra.

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 finitely generated. In general, we need to observe that the image of $V \to \operatorname{Hom}_{\text{\(R\)-mod}}(V^*, R[x])$ lies in the subalgebra of finite-rank homomorphisms (i.e., those whose image lies in the submodule generated by $\{1, \dotsc, x^n\}$ for some $n$).

$\newcommand\T{\mathrm T}$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, 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.

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.

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 finitely generated. In general, we need to observe that the image of $V \to \operatorname{Hom}_{\text{\(R\)-mod}}(V^*, R[x])$ lies in the subalgebra of finite-rank homomorphisms (i.e., those whose image lies in the submodule generated by $\{1, \dotsc, x^n\}$ for some $n$).