Most of the time it's going to be limited by the actual I/O speed of the disk. Even though the new ATA interfaces can do 133MB/s, the disks themselves can't keep up with that datarate.
I honestly don't know what the throughput is on a new drive on ATA/133 (and I don't want to wipe my main disks to find out). 20MB/s used to be a decent average, but the newer stuff is supposed to improve more over the medium stuff (as opposed to the medium stuff, which made almost no real-world difference over the old stuff). Let's say 30MB/s.
120,000MB / 30MB/s = 4000s per pass.
4000s x 7 passes (DoD standard) = 28,000s = 466.67m = 7.78h
So to do a full 7-pass DoD wipe on the whole drive, figure 8 hours if you're getting 30MB/s. If you're getting 60MB/s, it would only take 4 hours. If you're getting 60MB/s and only doing the short DoD (3 passes) it should only take a couple hours. If you're only getting 20MB/s, the original example will take about 12 hours instead of 8. If you could get the full 133MB/s the whole time, it should be under 2 hours instead of 8.
So the answer depends on how fast your drive is and how much wiping you're actually doing.
I have a 2 Western Digital 80 GB, 8 MB buffer hard drives, a Pentium 4 (2.0 GHz) with a SiS 650 chipset and it takes around 5 1/2 hours using the DOD method. The wipe does both drives at the same time. But I must use Boot and Nuke v. 1.0.2. (The older version did not work that fast.)