TL;DR

I did some preliminary checks to find that there were only 3 "random values" that are selected. Once you have a sample of the 3, you can XOR them with a known plaintext to get the seeds. Then you can sample the next 3 "random values", XOR them with the seed, then get the 3 candidates for the next letter.

Code

from pwn import *

octets = [0x77, 0x10, 0x2]

def sample(prefix):
    found_octets = set()
    while len(found_octets) < 3:
        io = remote("challenges.ctfd.io", 30468)
        io.sendlineafter("flag]>", prefix)
        io.recvuntil('unknown IPv4 address I have...')
        io.recvline()
        io.recvline()
        sample = io.recvline().strip()
        assert b"0x" in sample
        found_octets.add(int(sample, 16))
        io.close()

    return found_octets


import sys
prefix = sys.argv[1]
print(prefix)
o = sample(prefix)
o2 = [hex(_) for _ in o]
print("Octets", o2)
cset = set()
hset = set()
for i in o:
    for j in octets:
        ij = (i ^ j)
        cset.add(chr(ij))
        hset.add(hex(ij))

print(cset)
#print(hset)

I'm using pwntools because it's easy to play with stdin and stdout with it.

The octets were discovered beacuse I knew the first several letters would be flag{. When I typed in f, it would give me one of these 3 outputs: [0x1b, 0x7c, 0x6e].

XOR-ing them with the hex value of l gave me the seeds: [0x77, 0x10, 0x2]

The sample() function will query the server until it can gather 3 unique outputs.

Then it will print out cset ("character set") which contains all unique characters.

It's generally up to you to guess which one is next. I didn't automate this part.

Flag

flag{f0ll0w_th3_whit3_r@bb1t}