jmc

Let $d$ be the degree of $f(x).$ Then the degree of $g(f(x))$ is $2d=4,$ so $d=2.$

Accordingly, let $f(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}g(f(x))& =g(a{x}^2+bx+c)\\ & =(a{x}^2+bx+c{)}^2-11(a{x}^2+bx+c)+30\\ & =a^2{x}^4+2ab{x}^3+(2ac+b^2-11a)x^2+(2bc-11b)x+c^2-11c+30.\end{array}$$Comparing coefficients, we get $$\begin{array}{rl}{a}^2& =1,\\ 2ab& =-14,\\ 2ac+b^2-11a& =62,\\ 2cb-11b& =-91,\\ c^2-11c+30& =42.\end{array}$$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 $\boxed{11}.$