Your formula is true if you take $\nu$ to be the Minkowski content, assuming that the Minkowski contents in the integral are finite. This is because the Minkowski content is defined as the derivative of the volume of the tubular neighborhood.
One condition ensuring that Minkowski contents are finite is that the distance function to $A$ has Clarke gradient bounded below in the $\epsilon$-neighborhood of $A$. It also ensures that Minkowski contents coincide with the Hausdorff measures of the boundaries.
Finally, although that's not the question, it is possible to bound the volume of the tubular neighborhood without any conditions based on $\epsilon$ covering numbers of $A$.