What is the average weight of a minimal spanning tree of $n$ randomly selected points in the unit cube? Suppose we pick $n$ random points in the unit cube in $\mathbb{R}_3$, $p_1=\left(x_1,y_1,z_1\right),$ $p_2=\left(x_2,y_2,z_2\right),$ etc. (So, $x_i,y_i,z_i$ are $3n$ uniformly distributed random variables between $0$ and $1$.) Let $\Gamma$ be a complete graph on these $n$ points, and weight each edge $\{p_i,p_j\}$ by $$w_{ij}=\sqrt{\left(x_i-x_j\right)^2+\left(y_i-y_j\right)^2+\left(z_i-z_j\right)^2}.$$

Question: What is the expected value of the total weight of a minimal spanning tree of $\Gamma$?

(Note: Here total weight means the sum of all edges in the minimal spanning tree.)
A peripheral request: The answer is probably a function of $n$, but I don't have the computing power or a good implementation of Kruskall's algorithm to suggest what this should look like.  If someone could run a simulation to generate this average over many $n$, it might help towards a solution to see this data.
 A: According to the Wikipedia article on the Euclidean minimum spanning tree (or EMST), the expected total weight is asymptotic to
$$
c(d)n^{\frac{d-1}{d}}\int_{\mathbb{R}^{d}}f(x)^{\frac{d-1}{d}}dx
$$
as $n\rightarrow\infty$, where $d$ is the dimensionality (here, $d=3$), $f(x)$ is the probability density of point selection (here, $f(x)=1$ inside the cube), and $c(d)$ depends only on the dimensionality.  For uniform selection within the unit cube, the expected size should satisfy
$$
E(n) \sim c_{3}n^{2/3}
$$
for some constant $c_{3}$.
Numerical experiments (for modest $n$, up to about $8000$) suggest that $c_{3}=0.65 \pm 0.01$.
A: If $n = 0$ or $n = 1$ the answer obviously is 0. If $n = 2$ we have
$$E\left((x_1 - x_2)^2\right) = E(x_1^2 - 2x_1x_2 + x_2^2) = E(x_1^2) - 2E(x_1)\cdot E(x_2) + E(x_2^2) \\= \frac13 - 2\frac12\cdot\frac12 + \frac13 = \frac16.$$
The same for $y$- and $z$-coordinates. So $E(w_{12}) = \sqrt{\frac16  + \frac 16 + \frac16} = \frac1{\sqrt2}$ and spanning tree contains the edge $\{\,1, 2\,\}$ only.
I see it is possible to consider several cases for $n = 3$, however for arbitrary $n$ I don't expect to get close form of the answer.
