**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.

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user Juan Bermejo Vega