OK, so I stumbled upon something entertaining:
Apparently, a Lisp system written by Guy L. Steele for the IBM 1130. It's not written how much "free space" you have, but on page 25 of the second PDF is said: "1130 LISP CAN BE ASSEMBLED FOR A CORE SIZE OF 16K OR 32K. USING 32K APPROXIMATELY QUINTUPLES THE AMOUNT OF FREE STORAGE AND FIXED-POINT NUMBER SPACE AVAILABLE."
That, essentially, gives us two equations, where x is the available space and y is the size used by the system:
x + y = 16K
2*(x + y) = y + 5x
which means, the system uses by default 12K.
Which, essentially, means, if you had a 16K machine, you had only 4K for your LISP programs. I have no idea how exactly that translates into cons cells in this specific implementation, but sure enough I think we are in the territory that can be reached by uLisp. :)