I'm not confident of this answer, but it seems worth presenting:
NIntegrate[2^3/((x - s)^2 + (y - t)^2 + (z - u)^2),
{x, 0, 1}, {y, 0, 1}, {z, 0, 1}, {s, 0, x}, {t, 0, y}, {u, 0, z},
AccuracyGoal -> 16] // Quiet
(* 5.63378 *)
The basic insight is that each "paired" integrations (e.g., $x$ and $s$; $y$ and $t$; $z$ and $u$) is over the unit square. Each such pair can be replaced by a symmetric integration over half that square (e.g., $0 < x < 1$ with $0 < s < x$). Thus there will be an undercounting of $2^3 = 8$.
One can extend this core idea to integrating each "pair" of variables over a smaller portion of their respective unit square and multiply the overall integral accordingly.
Anyway... perhaps this will point others in the proper direction.
I now believe my answer is correct.
Here is why: Make a change of variables: $a = x - s$; $b = y - t$ and $c = z - u$. Then the integral over the unit square defined by $x$ and $s$ can be recast as a one-dimensional integral over the new variable $0 \le a \le 1$. However, there is a scaling factor that is needed which you can see if you draw the unit square and consider symmetry. There is hence a factor of $2 (1 - a)$ and likewise for the $b$ and $c$ variables.
NIntegrate[(8 (1 - a) (1 - b) (1 - c))/(a^2 + b^2 + c^2),
{a, 0, 1}, {b, 0, 1}, {c, 0, 1},
AccuracyGoal -> 16]
(* 5.63371 *)
If someone has the time, he or she could go one step further and replace the integration over the cube defined by $a$, $b$ and $c$ to a one-dimensional integration over a variable $q$ that goes on the diagonal of that cube, from $(a,b,c) = (0,0,0)$ to $(1,1,1)$. Be careful with scaling and normalization. If we're lucky, we might get a closed-form answer (presumably involving inverse trigonometric functions). Here's a figure illustrating the new variable $q$ in red:

Wouldn't that be cool? A six-dimensional numerical integration replaced by a one-dimensional symbolic integration?!
Integrate[1/((x - s)^2 + (y - t)^2 + (z - u)^2), x]
evaluates quickly butIntegrate[1/((x - s)^2 + (y - t)^2 + (z - u)^2), {x,0,1}]
can be done by hand before Mathematica finishes. $\endgroup$ – IPoiler Nov 5 '15 at 0:35