Here's a way to do it 48 edges.
A regular tetrahedron has four vertices each connected to the three other vertices by an edge of length one. Take four of these tetrahedra, and place four translated copied so that their centers themselves form a regular tetrahedron of edge length $1$.
Then, the corresponding vertices of each tetrahedron themselves form a regular tetrahedron with edge length $1$. So, each vertex is one away from $6$ other vertices -- three in its own tetrahedron, and the three corresponding points in other tetrahedra. This is a solution with $48$ edges -- each of $16$ vertices connects to $6$ other vertices, and $(16\times 6)/2 = 48$. If the original tetrahedron is rotated randomly, no edges cross with probability $1$
Mathematically, we've taken the Cartesian graph product $T\times T$, whose vertices correspond to pairs $(v,w)$ of vertices of $T$, and two vertices are connected by an edge, written $(v,w) \leftrightarrow (v',w')$, if either $v=v'$ and $w \leftrightarrow w'$, or $w=w'$ and $v \leftrightarrow v'$ in $T$. We've realized it as a unit distance graph in 3D as follows by taking the vector sum $v+w$ corresponding to each pair of vertices $(v,w)$, with a different tetrahedron embedding for $v$ and $w$.
This type of construction lets us create a unit distance graph with arbitrary many edges per vertex. If we have a solution for $n_1$ edges per vertex using $V_1$ vertices and likewise for $n_2$ and $V_2$, then taking the Cartesian product gives a solution for $n_1 + n_2$ edges per vertex that uses $V_1 V_2$ vertices. Note that minimizing vertices is equivalent to minimizing edges due to the relation $E=nV/2$.
