Let $K$ be an open bounded convex subset of $\mathbb R^n$ containing the origin and $\lvert x\rvert$ be the euclidean norm of $x$. We define the quantities $$ \begin{split} &\lVert x\rVert=\inf\left\{\lambda >0\,\middle|\,\frac{x}{\lambda}\in K\right\}\\ &\lVert x\rVert_*=\sup\left\{x\cdot y\,\middle|\, y\in K\right\} \end{split} \quad x\in \mathbb R^n $$
For every open bounded set $E$ with $C^\infty$ boundary, we define the function $$ \varphi: x\in\mathbb R^n\mapsto\begin{cases} d(x, \partial E) & \text{ if }x\notin E\\ -d(x, \partial E) &\text{ if }x\in E \end{cases} $$ where $d(x, \partial E)=\inf\{\lVert x-y\rVert | y\in\partial E\}$. I've been able to prove that, for every $\epsilon >0$ $$ \frac{\lvert E+\epsilon K\rvert-\lvert E\rvert}{\epsilon}=\frac{1}{\epsilon}\int_0^\epsilon dr\int_{\{\varphi = r\}}\left\lVert\frac{\nabla\varphi}{\lvert\nabla\varphi\rvert}\right\rVert_*d\mathcal{H}^{n-1} $$ where $\lvert E\rvert$ is the Lebesgue measure of $E$.
Now let $\nu_E$ the outer normal of $E$: I should prove that $$ \lim_{\epsilon\to 0^+}\frac{\lvert E+\epsilon K\rvert-\lvert E\rvert}{\epsilon}=\int_{\partial E}\lVert\nu_E\rVert_*d\mathcal{H}^{n-1} $$ but I don't know hot to recognize it because $K$ has only a Lipschitz boundary so $\varphi$ is only a Lipschitz function and $\nabla \varphi$ is only defined almost everywhere respect to Lebesgue measure. Any hint in order to prove it?
This paper answer to your question: https://arxiv.org/pdf/1203.5190.pdf. You do not need to take $E$ smooth: it suffices of finite perimeter.