This is the relationship between density operator and Bloch vector: $$\rho= \frac{1}{2}({\bf{\hat{1}}}+{\bf{b}}.\boldsymbol{\hat{\sigma}})$$
We define the NOT Universal Operator in the following way:
$$U: {\bf{b}}\to -{\bf{b}}/3$$
My question is - How does NOT Universal Operator act on the elements of the computational basis: $|0\rangle, |1\rangle$?
Note As it has been said in the comments, this definition of Universal-NOT gate seems to differ from others discussed in other posts [1]. This answer uses the definition proposed by the OP, i.e.
$$\rho=\frac{1}{2}(I+\vec{b}\cdot\vec{\sigma}) \quad \longrightarrow \quad U(\rho)=\frac{1}{2}\left(I-\frac{1}{3}\vec{b}\cdot\vec{\sigma}\right)$$ Where I use the symbol $I$ for the identity matrix to avoid confusion with 1 and $\vec{b}\in \mathbb{R}^3$ denotes the Bloch vector.
We write the density matrices of the computational basis states explicitly: $$ \rho_a=|a\rangle\langle a |= \frac{1}{2}(I+\vec{b}_a\cdot\vec{\sigma}),$$ where $a\in\{0,1\}$. Expanding this expression readily yields the vectors $b_a$: $$\vec{b}_a=(0,0,\pm1).$$ Applying your definition of $U$ to these density operators, the action of the operator on basis states can be obtained directly: $$ U(|0\rangle\langle 0|)= \frac{1}{2}(I- \frac{1}{3}\sigma_z)=\frac{1}{3}|0\rangle\langle 0| +\frac{2}{3}|1\rangle\langle 1 |$$ $$ U(|1\rangle\langle 1|)= \frac{1}{2}(I+ \frac{1}{3}\sigma_z)=\frac{2}{3}|0\rangle\langle 0| +\frac{1}{3}|1\rangle\langle 1 |$$ We can observe that, because the factor $1/3$ that "damps" the Bloch vector, pure basis states evolve into mixed states; notice that, intuitively, states get closer to the totally mixed state $I/2$ if you make the Bloch vector $\vec{b}$ go to zero.
