Probably not much. The most potent of the 4-halide 2C's is 2C-I, which is the bulkiest, but the I atom is the least electronegative of the common halogens. 2C-P is the most potent of the 4-alkyl 2C's - and the alkyl chain is pretty bulky, maybe twice that of an I atom all up.... but it's not very electronegative at all. See where i'm going with this?
Well, it would appear for high activity, you want bulk & electronegativity.
F atom is a bit smaller than a hydrogen. So, if we have fluorines all over our alkyl substituent at -4, as many as we can get, we should get great potency - and judging by our existing SAR data, a propyl would be our best choice.
AFAIK, these have not been synthesized yet, let alone human bioassayed, unfortunately.
So the answer to your question is - I don't know, but I suspect the 2 position would be best but can't figure out how to explain why :/ - but there's no reason you have to have them all at one position - theoretically, you could have 9 (!) fluoro subs on your propyl sub!
DOTFP would be crazy potent, haha. I for one will not be eating any however

, I don't like hour and hours and hours of neverending tripping.... but that's me. Each to their own.
3-Bromo-1,1,1-trifluoropropane is avaliable, so upon reaction with 2C-I you should get DO(3,3,3)TFP (all 3 fluoros at the end of the propyl sub)... well, once you seperate out the products, it'll be a messy seperation. That's just an example though, there are others.