test.py

import re
import execjs

content = '''
(String.fromCharCode
        (
            (010 * 12 + 6),
            (06 * (0x1 * (1 * 0xa + 6) + 1) + 12),
            (4 * 27 + 3),
            (01 * 0x3b + 50),
            (1 * 0x34 + 15),
            (1 * (1 * (3 * ((0x1 * 8 + 7) * 1 + 0) + 8) + 24) + 27),
            (0x1 * (2 * 0x25 + 7) + 16),
            (1 * 0112 + 40),
            (1 * 0x2c + 23),
            (0x3 * 042 + 9),
            (1 * ((05 * 4 + 1) * 03 + 0) + 37),
            (0x2 * (1 * 0x1f + 4) + 31)
        ))
        String.fromCharCode
        (
            (010 * 12 + 6),
            (06 * (0x1 * (1 * 0xa + 6) + 1) + 12),
            (4 * 27 + 3),
            (01 * 0x3b + 50),
            (1 * 0x34 + 15),
            (1 * (1 * (3 * ((0x1 * 8 + 7) * 1 + 0) + 8) + 24) + 27),
            (0x1 * (2 * 0x25 + 7) + 16),
            (1 * 0112 + 40),
            (1 * 0x2c + 23),
            (0x3 * 042 + 9),
            (1 * ((05 * 4 + 1) * 03 + 0) + 37),
            (0x2 * (1 * 0x1f + 4) + 31)
        )
'''
# soup = re.findall(r"\((.+?)\)", content, re.MULTILINE)
# print(soup)

looptime = len(re.findall(r"String.fromCharCode", content))

pepper = "String.fromCharCode"
text = content

for _ in range(looptime):
    num = text.find(pepper) + len(pepper)
    text = text[num:]

    soup = list(text)
    soup.append(' ') # Just to be sure
    i = 0
    first = int()
    last = int()
    for idx, vegetable in enumerate(soup):
        o = i
        if vegetable == '(':
            i += 1
        if vegetable == ')':
            i -= 1
        if i == 1 and o == 0: # First Opened Bracket
            first = idx
        if i == 0 and o == 1: # Last Closed Bracket
            break
    last = idx + 1
    content = content.replace(f"{pepper}{text[:first]}{text[first:last]}", execjs.eval(f"{pepper}{text[first:last]}"), 1)
    text = text[last:]

print(content)