Well, I'm not sure if this will be helpful or not, but this reminds me of finding isomers in organic chemistry...so maybe we could look at this with a similar systematic method as finding and naming of isomers:
Consider the longest straight "chain", then the next longest chain, and so on...
For n cubes, there is 1 arrangement with the longest chain of n cubes.
There are (n-1)/2 for odd n and n/2 for even n arrangements with the longest chain of n-1 cubes. (You can put the nth cube on a face of any of the other cubes except the end faces of the end cubes, since that would just be the n cube chain, and putting it on the 1st cube is the same as putting it on the last cube, the 2nd is the same as the 2nd to last, etc. This is equivalent to the rule in OChem that you start counting at the end that will give you the smallest numbers)
For the longest chain being n-2, you have different options. You can either attach a 2 cube "side-chain" or two 1 cube side-chains to the base chain. There are (n-2)/2 for even n or (n-1)/2 for odd n ways of attaching the 2 cube side-chain. For the two 1 cube chains, you can start out putting 1 cube on the first cube in the base chain, then put the second 1 cube side chain either 90 degrees or 180 degrees from it on each of the n-2 cubes in the base chain, for 2(n-2) arrangements. Then move the first cube side-chain and repeat the process until you get to the center cube for another 2(n-3)+ 2(n-4)+...2(n-n/2) for even n or 2(n-(n-1)/2) for odd n arrangements.
Okay, I can see that this is getting complicated really quickly, and as mentioned by others, there is the issue of "chirality" to consider when the side chains are different lengths...but, I dunno, this might be a start or help in some way...I'll keep thinking about it (whether I want to or not...I'm one of those ppl who can't stop thinking about a problem once they get in into their head...X\)