Can someone help me with the amount of water grain typically absorbs? Anything I have forgotten?
I've always used 0.1 Gal / LB of grain, though I've heard lower numbers if you BIAB and squeeze. I also use a similar number (0.75 G / oz) for pellet hop absorption, but again this depends if you're just chucking them in, using a spider/bag and squeezing a little etc)
Like others have mentioned, the answer really depends on your brewing setup/process. I try to calculate backwards from my end result going into the keg to make sure I've got a good fill with not too much waste. As a rough example, if I:
Target 5 G into the keg
+0.5 G for dry hop absorption + fermenter losses = 5.5 G needed in the fermenter
5.5 Gal + 0.25 G cooling loss (shrinkage!) + 0.5 G hop / trub / chiller losses = 6.25 G needed for my post-boil volume
Pre-boil volume is essentially your post-boil volume + your boil off rate (and possibly + losses if transferring). If the recipe calls for a 60 min boil, and I boil off 0.75 G / hour I get:
6.25 G + 0.75 G = 7.0 G pre-boil target
From there I work in the grain absorption (lets say it's 10lb of grains @ 0.1 Gal / LB for easy math)
7 G + 1 G absorption = 8 G total water needed
For a 10 lb grain bill, if you target a mash thickness of 1.3 quarts / LB, it's 13 quarts or 3.25 G strike water
Total water (8 G) - Strike water (3.25 G) = 4.75 G sparge water
Almost all of these variables (boil rates, absorption, losses etc) heavily depend on the process and equipment you're using to dial in your numbers. I agree with IslandLizard, the calculators are your friends especially once you know what your typical losses and rates are.
Hope this helps!