Your first word is 7 digits your second is 3, so clearly one is stronger than the other. "nom" is not in the diceware set, which helps a little, but it isn't so uncommon to be in a search dictionary. The numbers are in the diceware set.

You're comparing 7700^3 against 7700^7. Your more secure password isn't any better than chickensandwichwafflesworkcraigcrossafrica, probably a lot less good because chicken, delicious and nom clearly correlate heavily and nomnomnom is almost one word really. 7700^7 is 1604852326685300000000000000 according to my calculator. If I assume 72 characters (52 letters, 10 numbers, 10 special characters) then I need a 15 character random password to beat it in terms of search space. Maybe this: }&X$0ueUo~ravx&.

Further, if you put numbers between your letters you are turning a search space of 7700 into 7710 or whatever. If you replace l with 1 and so on, you are surely turning 7700 into 7700*(number of replacement options and combinations thereof). So mathematically, I would think that replacing e with 3, a with @ would actually be a stronger encoding that what you suggest.