Late answer, but since $\nu$ can be shown to be close to a measure that satisfies a dimension-free LSI, we can obtain a dimension-free constant using the Holly-Stroock pertubation lemma (eg see Proposition 5.1.6 in Bakry's book).

For example, if we define the measure $\mu \propto e^{-g}$ with $g(x) = f(x) + \frac{\mu}{4} \|x\|^2 \mathbb{1}_{\|x\| \leq 1}$, then we have that $\lambda_{\min}(\nabla^2 g) \geq \frac{\mu}{2}$ and thus, $\mu$ satisfies an LSI with constant $\frac{\mu}{2}$. Furthermore, since $|f-g| \leq \frac{\mu}{4}$, it follows from the pertubation lemma that $\mu$ satisfies an LSI with constant $\frac{\mu}{2} e^{-\mu/2}$. The downside of this approach is that it is a weak estimate in the setting where $\mu$ is large – this can be partially remedied by fiddling with the function $g$.

Late answer, but since $\nu$ can be shown to be close to a measure that satisfies a dimension-free LSI, we can obtain a dimension-free constant using the Holly-Stroock perturbation lemma (eg see Proposition 5.1.6 in Bakry's book).

For example, if we define the measure $\mu \propto e^{-g}$ with $g(x) = f(x) + \frac{\mu}{4} \|x\|^2 \mathbb{1}_{\|x\| \leq 1}$, then we have that $\lambda_{\min}(\nabla^2 g) \geq \frac{\mu}{2}$ and thus, $\mu$ satisfies an LSI with constant $\frac{\mu}{2}$. Furthermore, since $|f-g| \leq \frac{\mu}{4}$, it follows from the perturbation lemma that $\mu$ satisfies an LSI with constant $\frac{\mu}{2} e^{-\mu/2}$. The downside of this approach is that it is a weak estimate in the setting where $\mu$ is large – this can be partially remedied by fiddling with the function $g$.