Let $ d $ be the degree of $ f(x) $. Then the degree of $ g(f(x)) $ is $ 2d = 4, $ so $ d = 2 $. Accordingly, let $ f(x) = ax^2 + bx + c $. Then \begin{align*} g(f(x)) &= g(ax^2 + bx + c) \\ &= (ax^2 + bx + c)^2 - 11(ax^2 + bx + c) + 30 \\ &= a^2 x^4 + 2abx^3 + (2ac + b^2 - 11a) x^2 + (2bc - 11b) x + c^2 - 11c + 30.\end{align*}Comparing coefficients, we get \begin{align*} a^2 &= 1, \\ 2ab &= -14, \\ 2ac + b^2 - 11a &= 62, \\ 2cb - 11b &= -91, \\ c^2 - 11c + 30 &= 42.\end{align*}From $ a^2 = -1, $ $ a = 1 $ or $ a = -1 $. If $ a = 1, $ then from the equation $ 2ab = -14, $ $ b = -7 $. Then from the equation $ 2cb - 11b = -91, $ $ c = 12 $. Note that $ (a,b,c) = (1,-7,12) $ satisfies all the equations. If $ a = -1, $ then from the equation $ 2ab = -14, $ $ b = 7 $. Then from the equation $ 2cb - 11b = -91, $ $ c = -1 $. Note that $ (a,b,c) = (-1,7,-1) $ satisfies all the equations. Therefore, the possible polynomials $ f(x) $ are $ x^2 - 7x + 12 $ and $ -x^2 + 7x - 1 $. Since \[x^2 - 7x + 12 + (-x^2 + 7x - 1) = 11\]for all $ x, $ the sum of all possible values of $ f(10^{100}) $ is $ \boxed{11} $.