For completeness, let me copy the essence of my answer from over at physics.SE:

The W state is not a stabilizer state - for a stabilizer state, the 1-site reduced density matrices must be maximally mixed or pure, which they aren't.

To cite from my answer from over at physics.SE: The W state is not a stabilizer state - for a stabilizer state, the 1-site reduced density matrices must be maximally mixed or pure, which they aren't.

Or, to phrase it without reduced density matrices: For a stabilizer state, if you measure $X$, $Y$, or $Z$ for any single qubit, the probability of getting either outcome is either $0$, $1/2$, or $1$. This is clearly not the case for a $Z$ measurement on the W state above, where the probability of getting $0$ is $1/3$.