Boundary of the intersection of two open sets in $\mathbb{R}^n$

This is not true generally unless $\overline{A\cap B}=\overline{A}\cap \overline{B}$. \begin{align} \partial (A\cap B)&= \overline{A\cap B}-(A\cap B)^{o} \\ &=(\overline{A}\cap \overline{B})-(A^{o}\cap B^{o}) \\ &=(\overline{A}\cap \overline{B})\cap(A^{o}\cap B^{o})^c \\ &=(\overline{A}\cap \overline{B})\cap(A^{o^c}\cup B^{o^c}) \\ &=(\overline{A}\cap \overline{B}\cap A^{o^c})\cup(\overline{A}\cap \overline{B}\cap B^{o^c}) \\ &=(\overline{A}\cap \partial B)\cup (\overline{B}\cap \partial A) \end{align} By this post, $\overline{A\cap B}=\overline{A}\cap \overline{B}$ implies discrete space. So this is impossible in $\Bbb{R}^n$. However, we can prove generally $$ \partial(A\cap B)\subset (\bar A \cap \partial B) \cup (\partial A \cap \bar B) $$ for it is always true that $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$. This can be done easily by replacing "$=$" with "$\subset $" at 2nd line of above proof and rest follows.

It does not hold. Consider for example $$A=\{x\in \mathbb{R}^n:|x|<1\}$$ and $$B=\{x\in \mathbb{R}^n:|x-(2,0,\dots,0)|<1\}.$$ Since $A\cap B=\varnothing$, $\partial(A\cap B)=\varnothing$, but the RHS in your formula is the set $\{(1,0,\dots,0)\}$.