Let's take the case $n=2$, rewrite the inequalities as $$a\le x\le b,\qquad c\le y\le d,\qquad x\ge y$$ We're given $a,b,c,d$ and want to count the number of pairs $x,y$. We assume $a\le b$ and $c\le d$, as otherwise the answer is zero. In fact, I'll assume $a,b,c,d$ distinct just to keep my calculations simple.
We have 6 cases:
$c\lt d\lt a\lt b$;
$c\lt a\lt b\lt d$;
$c\lt a\lt b\lt d$;
$a\lt c\lt d\lt b$;
$a\lt c\lt b\lt d$;
$a\lt b\lt c\lt d$.
In the first case, the answer is $$(b-a+1)(d-c+1)$$
In the second case, $$(a-c+1)+(a-c+2)+\cdots+(d-c)+(b-d+1)(d-c+1)$$
Third case case, $$(a-c+1)+(a-c+2)+\cdots+(b-c+1)$$
Fourth case, $$1+2+\cdots+(d-c)+(d-c+1)(b-d+1)$$
Fifth case, $$1+2+\cdots+(b-c+1)$$
Sixth case, zero.
All the $\cdots$ indicate the terms go up by one, so it shouldn't be hard to evaluate them all as sums of arithmetic progressions. My point is, that even in the case $n=2$ you get a slew of different answers, depending on the order relations among the parameters. I really don't think there's going to be a fully general yet useful answer to the question.