Here’s a partial answer to get you started. If I understand this correctly, you have in general $$F(a_1,\dots,a_n) = \sum\limits_{x_1=0}^{a_1}\sum\limits_{x_2=0}^{a_2}\dots\sum\limits_{x_n=0}^{a_n}F'(x_1,\dots,x_n).$$ For each $S \subseteq \{1,\dots,n\}$ let $F^S(a_1,\dots,a_n) = F(b_1,\dots,b_n)$, where $b_k = a_k-1$ if $k \in S$, and $b_k = a_k$ otherwise.
Now suppose that $0 \le c_k \le a_k$ for $k=1,\dots,n$, and let $S = \left\{k \in \{1,\dots,n\}: c_k < a_k \right\}$. The term $F'(c_1,\dots,c_n)$ is included in the sum $F^T(a_1,\dots,a_n)$ iff $T \subseteq S$. Let $$G(a_1,\dots,a_n) = \sum\limits_{k=1}^n (-1)^{k+1}\sum\limits_{|T|=k}F^T(a_1,\dots,a_n),$$ where it’s understood that $T$ in the inner sum ranges over subsets of $\{1,\dots,n\}$. The term $F'(c_1,\dots,c_n)$ is counted $\sum\limits_{k=1}^{|S|}(-1)^{k+1}\binom{|S|}{k}$ times in this sum. But $\sum\limits_{k=1}^{|S|}(-1)^{k+1}\binom{|S|}{k} = (-1)\sum\limits_{k=1}^{|S|}(-1)^k \binom{|S|}{k} = (-1)\left(\sum\limits_{k=0}^{|S|}(-1)^k \binom{|S|}{k} - 1\right) = 1$, so $$F(a_1,\dots,a_n) = F'(a_1,\dots,a_n)+G(a_1,\dots,a_n),$$ and $$\begin{align*}
F'(a_1,\dots,a_n) &= F(a_1,\dots,a_n) - \sum\limits_{k=1}^n (-1)^{k+1}\sum\limits_{|T|=k}F^T(a_1,\dots,a_n) \\
&= F(a_1,\dots,a_n) + \sum\limits_{k=1}^n (-1)^k\sum\limits_{|T|=k}F^T(a_1,\dots,a_n).
\end{align*}$$ When $n=2$ this essentially reduces to your formula $$F(x,y) = F'(x,y) + F(x-1,y) + F(x,y-1) - F(x-1,y-1).$$
There are adjustments to be made when one or more of the $a_k$ are $0$, but apart from that, this recovers $F'$ from $F$, and from that it shouldn’t be too hard to get the interval sums that you want. If I have time later (and no one beats me to it) I’ll work those out as well.