A triple summation with a restriction!

We'll solve the problem using inclusion-exclusion principle.
We first calculate the value without the condition that $i,j,k$ are distinct. $\sum\frac{1}{2^i2^j2^k} = \sum 2\times\frac{1}{2^i 2^j} = \sum 2\times 2\times\frac{1}{2^i} = 8$
Now we need to remove the extra value in our sum by removing all those cases which do not satisfy our condition. We consider the case that two of our variables are equal without any restriction on the third variable.
$\sum \frac{1}{2^i 2^j 2^k} = \sum \frac{1}{2^{2i} 2^k}= \sum 2\times \frac{1}{2^{2i}} = 2.\frac{4}{3} =\frac{8}{3}$
There will be $3$ such cases, namely $i=j, j=k \text{ and } k=i$ . So we need to subtract three times this value from our previously calculated value. But doing so, we've also removed the case when all three variables are equal thrice. Since we want to remove this case only once, we add twice its value. $\sum \frac{1}{2^i 2^j 2^k} = \sum \frac{1}{2^3i} = \frac{8}{7} \\ 8 - 3\times\frac{8}{3}+ 2\times\frac{8}{7} = \frac{16}{7}$
Since $16,7$ are co-prime, the answer will be $\boxed{23}$ .

This is something of a special case of Euler's work on the generating function of the partition function $P$ .

Let us let $S(n)$ be the sum $S(n) \; =\; \sum_{{a_1,a_2,\ldots,a_n \ge0} \atop {a_1,a_2,\ldots,a_n \text{ distinct}}} 2^{-(a_1+a_2+\cdots+a_n)} \;= \; n! \sum_{0 \le a_1 < a_2 < \cdots < a_n} 2^{-(a_1+a_2+\cdots+a_n)} \;,$ Then $S(n) \; = \; n! \sum_{N \ge \frac12n(n-1)} \big|A(N,n)\big| 2^{-N} \;,$ where $A(N,n)$ is the set $A(N,n) \; = \; \left\{ \mathbf{a} \in \mathbb{N}\cup\{0\}^n \; \Big| \; a_1 < a_2 < \cdots < a_n \;, \; \sum_{j=1}^n a_j \,=\, N \right\} \;, \; N \ge \tfrac12n(n-1) \;.$ If we also consider the set $B(L,n) \; = \; \left\{ \mathbf{b} \in \mathbb{N}\cup\{0\}^n \; \Big| \; \sum_{j=1}^n (n+1-j)b_j \,=\, L\right\}\;, \qquad L \ge 0 \;,$ then the mapping $X \,:\, B\big(N-\tfrac12n(n-1),n\big) \,\to\, A(N,n)$ given by the formula $(X\mathbf{b})_j \; = \; j-1 + \sum_{r=1}^j b_r \qquad 1 \le j \le n \;, \qquad \qquad \mathbf{b} \in B\big(N-\tfrac12n(n-1),n\big)$ is a bijection for any $N \ge \tfrac12n(n-1)$ , and hence $\big|A(N,n)\big| \,=\, \big|B(\big(N-\tfrac12n(n-1),n\big)\big|$ .

But $\big|B(L,n)\big|$ is the coefficient of $t^L$ in the series expansion of the product $\prod_{r=1}^n \big(1 - t^r\big)^{-1} \;,$ and hence $\big|A(N,n)\big|$ is the coefficient of $t^N$ in the series expansion of the product $Q_n(t) \; = \; t^{\frac12n(n-1)}\prod_{r=1}^n \big(1 - t^r\big)^{-1} \; = \; \prod_{r=1}^n \left(\frac{t^{r-1}}{1-t^r}\right) \;.$ Thus it follows that $S(n) \; = \; n! Q_n(\tfrac12) \; =\; n! \prod_{r=1}^n \frac{2}{2^r-1} \; = \; n!2^n \left(\prod_{r=1}^n (2^r-1)\right)^{-1} \;.$ In this case, we have $S(3) \; =\; \frac{3! \times 2^3}{1\times3\times7} \; =\; \frac{16}{7} \;,$ making the answer $16+7 = \boxed{23}$ .