The free stress-energy tensor can be defined as

$$T^{\mu\nu}_{\rm free} = \int f(x^\mu,p_\mu) p^\mu p^\nu \frac{d^3 p}{p_0 m^2}$$

where $f$ is the distribution function, and I assume massive particles of mass $m$. The interaction stress-energy tensor can then be chosen so that $T^{\mu\nu}_{\;\;;\nu}$ is equivalent to the second moment of the collisional Boltzmann equation

$${f,H_{\rm free}} = \frac{\delta f}{\delta t}\Big|_{\rm coll.}$$

Generally, this is not possible. However, it is generically possible to obtain a volume-averaged version of the second moment that offers an asymptotic description of the dynamics in terms of an averaged stress-energy tensor $\bar{T}^{\mu\nu}$. This amounts to a free stress energy tensor with additional viscosity terms that scale as the gradients of stress energy times kinematic viscosity, $\sim \nu T^{\mu\nu}_{{\rm free};\kappa}$. However, the volume averaging itself adds terms

$$\bar{T^{\mu\nu}}_{\rm free} \equiv \frac{1}{\delta V}\int_{\delta V} T^{\mu\nu}_{\rm free} d^3 x = T^{\mu\nu}_{\rm free} + \propto T^{\mu\nu}_{\rm free;\kappa} \delta V^{1/3} + \propto T^{\mu\nu}_{\rm free;\kappa\gamma} \delta V^{2/3} + ...$$

In other words, the averaging volume itself adds kinematic viscosity, or $\Delta \nu/\Delta V^{1/3} \sim C$, where $C$ is a dimensionless geometrical-type constant. By going quantum, one can assume that $\delta V^{1/3}$ needs to be about as large as the De Broglie wavelength or smaller for convergence (note I assume massive particles), and assuming an ideal-gas-type behaviour, we have $\delta V^{1/3} \sim h/ \sqrt{m k_{\rm B} T}$, so I get

$$\Delta \nu \sqrt{m k_{\rm B} T} \lesssim h$$

This is really nothing like your relation. However, it can perhaps be useful for you to see that what kind of relation you get depends on the physical context you cite.

I think you might get interesting stuff from a gradient-type expansion of a conformal quantum fluid rather than massive non-relativistic gases as considered in this answer.