I think the answer is no. Indeed, consider the case when $n=2$ and, over the rectangle $R:=[-1,1]\times[-h,h]$, the function $f$ is the pointwise maximum of the set of all affine functions $g$ such that $g(0,0)\le-1$, $g(0,h)\le-1+2h$, $g(0,-h)\le-1+2h$, $g(1,h)\le0$, $g(1,-h)\le0$, $g(-1,h)\le0$, and $g(-1,-h)\le0$, where $h>0$ is small enough, with $f$ appropriately extended outside the rectangle $R$. 

Explicitly, for $h=1/10$,
$$f(x,y)=\max\left(| x| -1,\frac{4 | x| }{5}+2 | y| -1\right)$$
for $(x,y)\in R$, with 
$$f(x,y)=\sup\{f(u,v)+p\cdot(x-u,y-v)\colon \\ 
(u,v)\in(-1,1)\times(-h,h), p\in\partial f(u,v)\}$$
for $(x,y)\in\mathbb R^2$, where $\cdot$ denotes the dot product.