There are a lot of open sets in $\mathbb{R}^n$ whose boundary has positive Lebesgue measure. I wouldn't be surprised if "most" open sets have boundaries with positive Lebesgue measure, where "most" might be referring to cardinality, or some topological or measure-theoretic size.
However, the open sets you can visualize are far more regular than the average open set, so it's not easy (if at all possible) to get a good mental picture of an open set whose boundary has positive Lebesgue measure. And the open sets one does analysis on, typically also are quite regular and have nice boundaries.
Examples of open sets whose boundary is not a null set are for example complements of a thick Cantor set in dimension $1$ (products where at least one factor is such in higher dimensions).
Somewhat similar, let $(r_k)_{k \in \mathbb{N}}$ be an enumeration of the points with rational coordinates, and let
$$U = \bigcup_{k\in \mathbb{N}} B_{\varepsilon_k}(r_k)$$
for a sequence $\varepsilon_k \searrow 0$ such that $\sum {\varepsilon_k}^n$ converges. Then you have a dense open set $U$ with finite Lebesgue measure, its boundary is its complement and has infinite Lebesgue measure.
Is the stronger statement: "$U\subset \mathbb{R}^n$ open $\Rightarrow \partial U$ has measure zero" true as well?
No. In 1902 Wiliam F. Osgood presented his construction of "A Jordan curve of positive Area".
That provides an open set $U \subset \mathbb{R}^2$ such that $\partial U$ - the Jordan curve of positive area - has positive Lebesgue measure. Barring better ideas, you can use a product of such an open set with a (hyper)cuboid to have higher-dimensional examples.
The complement of a fat Cantor set provides a one-dimensional example.
Best Answer
Here is a sufficient geometric condition for a general set $E$ to have measure zero: for every $x\in E$ there is $c>0$ such that for all sufficiently small $r>0$ the $r$-neighborhood of $x$ contains a ball of radius $cr$ that is disjoint from $E$. This is a weak form of condition known as porosity. To see that it implies having measure zero, use the Lebesgue density theorem.
Porosity is easy to verify for the boundary of a given open set: it suffices to find, for every boundary point, a subset in the shape of a cone (possible twisted) with a vertex at that point. The shape and size are allowed to depend on the point. Smooth, Lipschitz and uniform domains are covered by this condition.