On the definition of (small) inductive dimension A regular topological space $X$ has inductive dimension smaller or equal to n if and only if:
($n=-1$) $X=\emptyset$;
($n>-1$) The space $X$ has a base of opens $\mathscr{B}$ such that, for all $B\in\mathscr{B}$,  $\delta B$ (the boundary of B) has inductive dimension smaller or equal to $n-1$.
Now.. if $X$ has inductive dimension smaller or equal to $n$, can we say something about the dimension of the boundary of a general open subset of $X$? 
 A: I would be surprised if there was a general result regarding boundaries of general open sets in spaces such that $ind(X)=n$. The only thing that comes to mind is the following:
Prop: If $X$ is a regular space such that $ind(X)=n<\infty$ then for each $m=-1,0,1,\ldots,n-1$ there is a closed subspace $C\subseteq X$ such that $ind(C)=m$.
In proving this it's actually a lot easier to use a different characterization of $ind$ in regular spaces.
Prop: A regular space $X$ satisfies the inequality $ind(X)\leq n\geq 0$ ($n<\infty$) if and only if for every point $x\in X$ and closed subset $B\subseteq X$ no containing $x$, there is some closed set $L\subseteq X$ such that $L\cap(B\cup\{x\})=\emptyset$ and $L$ separates $X$ between $x$ and $B$ and $ind(L)\leq n-1$.
When we say that $L$ separates $X$ between $x$ and $B$ we mean that there are disjoint open sets $U,V\subseteq X$ such that $x\in U$, $B\subseteq V$, and $X\setminus L=U\cup V$. Depending on the book you are looking at the set $L$ can be called a separator between $x$ and $B$ or a partition between $x$ and $B$.
With this particular characterization in hand it's a bit easier to prove that first proposition I stated, because if $ind(X)=n<\infty$ then there must be some $x\in X$ and some closed set $B\subseteq X$ not containing $x$ such that for every partition $L$ between $x$ and $B$ one has $ind(L)=n-1$ or $ind(L)=n$. Then you can proceed to the subspace $L$ and find subspaces of even smaller dimension.
I suppose one general result that could be stated is that if $U\subseteq\mathbb{R}^{n}$ is an open set then one necessarily has that $ind(\partial U)\leq n-1$ as a set in $\mathbb{R}^{n}$ has dimension $n$ if and only if it has nonempty interior. 
It is also pretty easy to find subspaces of $\mathbb{R}^{n}$ with open subspaces in which boundaries of open sets have to have arbitrarily small dimension. Just attach lower dimensional cubes to higher dimensional cubes.
