In the quadratic case, it does. Given an irreducible quadratic polynomial $f(x)=ax^2+bx+c$, the discriminant of the quadratic number field $\frac{\mathbb{Q}[x]}{f(x)}$ is $\operatorname{sqf}(d)$ or $4\cdot \operatorname{sqf}(d)$, depending on if $d \equiv 1 \mod 4$, where $d=b^2-4ac$ is the polynomial discriminant of $f(x)$, and $\operatorname{sqf}$ is the squarefree part.

One can reduce this to the case of local fields, in which one can ask a similar question. Fix a positive integer $n$. Given a separable polynomial $f \in \mathbb{Q}_p[x]$ of degree $n$, does the discriminant of $f$ determine the valuation of the discriminant ideal of the associated extension? What about in the case where $p \nmid n$?

For the question over local fields, one should note that the dependence on the discriminant of $f$ does *not* factor through the valuation of the discriminant of $f$.