Does this hold in general? I know the following holds when $n=1$. What about the case of $n \geq 2$?
Let $O \subsetneq \mathbb{R}^n$ be a nonempty open set.
There exists $C, \delta >0$ such that $|\{x \in O \mid \operatorname{dist}(x, \mathbb{R} \setminus O) < \varepsilon\}| \geq C {\varepsilon}^n$ for all $0<\varepsilon < \delta$.
 A: YES. Assume $S\ne \Bbb R^n.$ 
(I). Terminology. Let $m(S)$ denote the $n$-dimensional measure of any  measurable  $S\subset \Bbb R^n.$ For $x\in \Bbb R^n$ and $r>0$ let $B(x,r)$ be the open ball of radius $r$ centered at $x$ . Let $r^n C_n$ be the volume (measure) of $B(x,r).$ Let $C=2^{-n}C_n.$ 
Let $O^c=\Bbb R^n$ \ $O.$ For brevity let $\{x\in O: d(x, O^c)< e\}=S(e).$
(II). Take $d>0$ small enough that there exists $x_0\in O$ with $d(x_0,O^c)=d.$
For any $e\in (0,d)$   take $x_1\in \{tx_0+(1-t)q:t\in [0,1]\}$ such that $d(x_1,q)=e/2.$ Note that $d(x_0,x_1)=d-e/2.$ 
Now $B(x_1,e/2)\subset O .$ Otherwise,  if $y\in B(x_1,e/2)\cap O^c$ then $$d(x_0,O^C)\leq d(x_0,y)\leq d(x_0,x_1)+d(x_1,y)=(d-e/2)+d(x_1,y)< (d-e/2)+e/2=d.$$
So  for any $x\in B(x_1,e/2)$ we have $x\in O$ and $$d(x,O^c)\leq d(x,q)\leq d(x,x_1)+d(x_1,q)<e/2+d(x_1,q)=e. $$
Therefore $B(x_1,e/2)\subset S(e).$ 
Therefore $m(S(e))\geq m(B(x_1,e/2)=Ce^n.$
Footnote. $U=O^c \cap \overline {B(x_0,1+d)}$ is compact  so some $q\in U$ satisfies $d(x_0,q)=d(x_0, O^c) .$
