So the question is really to 
>! count the maximal number of cubelets in transparent mode.  
>! If there are $n$ of them, then the power needed is $4\times1000+n\times100+(121-n)\times10=5210+90\times n$ Watts.

A lower and upper bound for 
>! this maximal number is 18 and 66 respectively.  
>!  
>! The lower bound comes from an actual setup where the interferometer cubes are in 4 corners of the large cube, so they are pairwise in opposite corners of a side of the cube. (See the blue cubelets on the picture in the next paragraph.) Any two of these have 3 cubelets between them, different pair of interferometer cubes on different sides of the cube, so those between them are all distinct cubelets, that's $6\times3=18$ in total.  
>!  
>! However, this case is special because of the laserlight passing through the edges of the cubelets.  
>! In a more usual case a laserlight which goes through two cubelets which only share an edge between them, it is done through one of their common neighbours.
>! Two cubelets in opposite corners of the cube can be reached through 11 other cubelets with a path in which consecutive cubelets share sides.
>! Four cubelets make 6 different pairs, hence $6\times11$ is a very rough upper bound - actually the 4 cubelets can not all be in opposing corners to each other, and the sets of cubelets between those pairs probably have some common elements, which should not be counted twice.

There are some parts of the question for which I ave a generic method.  
>! How many cubelets need to be transparent between two interferometer cubelets, whose coordinates are $[x_1,y_1,z_1]$ and $[x_2,y_2,z_2]$?  
>! Noting the absolute distances for each coordinate with $[d_x,d_y,d_z]=[|x_1-x_2|,|y_1-y_2|,|z_1-z_2|]$ the answer for this is $d_x+d_y+d_z-\gcd(d_x+1,d_y+1)-\gcd(d_x+1,d_z+1)-\gcd(d_y+1,d_z+1)+2\gcd(d_x+1,d_y+1,d_z+1)$, where $gcd(\cdot)$ notes the greatest common divisor.

My guess is that the solution
>! for $n$ is exactly in the middle of these bounds: 42. That would have a power need of 8990W, that is approximately 9kW.  
>! I think a setup which needs 42 transparent cubelets can easily be shown: just put a slight rotation on the maximal tetrahedron fitting in the cube, so its corners are neighbouring the corners of the cube. Here it is on a picture to help understanding:  
>! (Blue cubelets are interferometers for the $n=18$ case. Brownish ones - and the fourth blue corner - produce around 40, I think, but I did not count precisely. Not yet.)  
>! [![a lower bound and a probably close to maximal setup][1]][1]  
>!  
>! But I might be totally wrong. It's just my mathematical intuition speaking.


  [1]: https://i.sstatic.net/pUJP0.png