I hate to bug you again, but I had trouble burning the bootloader to my 1284P, so I attempted some troubleshooting and read a bunch of other things.
Since lock bits are not clear in my mind, I will simply state my findings, without attempting to comment. So here it is: maniacbug has the lock bits as 0x0F instead of 0x2F. I was getting a verification error when attempting to burn the bootloader with 0x2F for the lock bit, but I was successful with 0x0F.
I will do some reading in an attempt to understand the lock bits, but I thought I should mention this. And I apologize for not specifying earlier that I am looking at boards.txt for 1.6
Edit: After spending some time reading the datasheet, I think this is what needs to happen: Unprogramming (1) bootloader lock bits BLB11 and BLB12 (0x3F) removes any restrictions accessing the boot loader, necessary to be able to write it to the chip. Programming (0) them both (0x0F) locks writing to the boot loader and reading from it as well. Only programming BLB11 (0x2F) just locks the bootloader for writing.
So, if I understand this correctly, both 0x0F and 0x2F lock the bootloader for write, which makes sense so that it doesn't get wiped out, but 0x2F allows the program to read from it.
In my case, both lock bit settings should have worked, as both are valid. I will go back to 0x2F and retry, maybe I messed something else up.
Edit (again - sorry): It worked with 0x2f, so it was me.
Sorry for wasting your time.