Answer: 151

Solution:

The base-sixteen representation of $g(n)$ needs a digit beyond $9$ exactly when $g(n) \ge 10.$ So we need the base-eight digit sum of $f(n)$ to be at least $10.$ Checking values in order, every number less than $31$ has base-eight digit sum at most $9,$ while $31 = 37_8$ has digit sum $10.$ Since the least $n$ achieving a given base-four digit sum increases with that sum, we want the least $n$ with $f(n) = 31.$

A base-four digit is at most $3,$ so a digit sum of $31$ requires at least $11$ digits, and the smallest $11$-digit choice is a leading $1$ followed by ten $3$s: $$\begin{aligned} N &= 13333333333_4 \\ &= 4^{10} + (4^{10} - 1) \\ &= 2 \cdot 4^{10} - 1 = 2097151. \end{aligned}$$

The remainder when $N = 2097151$ is divided by $1000$ is $151.$