For a centered Gaussian measure $\mu$ on a Hilbert space $X$, it is known that
$$\int_X \|x\|^2 \mu(dx) = tr(Q)$$ where $Q$ is the covariance operator. Is there a similar version for Gaussian measures in Banach spaces? I know that I can bound it from above by Fernique's theorem but I'm interested in calculating this explicitly.
There is no explicit expression for $\int_X \|x\|^2 \mu(dx)$ in general, even if $X=\mathbb R^d$ with $d\in\{2,3,\dots\}$ and $\mu$ is the standard normal distribution on $\mathbb R^d$.
Indeed, any norm on $X=\mathbb R^d$ is characterized by the corresponding unit ball $B:=\{x\in X\colon\|x\|\le1\}$. More specifically, $\|\cdot\|=M_B$, where $M_B$ is the Minkowski functional for $B$, defined by the formula $$M_B(x):=\inf\{t\in(0,\infty)\colon x/t\in B\}$$ for $x\in X$. Moreover, a set $C\subset X=\mathbb R^d$ is the unit ball for some norm on $X$ iff $C$ is compact, convex, symmetric (i.e., $-C=C$), and absorbing (i.e., $\bigcup_{n>0}nC=X$); let us call such a set a Minkowski set. Equivalently, a set $C\subset X=\mathbb R^d$ is the unit ball for some norm on $X$ iff $C=D-D$ for some compact convex set $D$ with nonempty interior. Thus, the set of all norms on $X=\mathbb R^d$ is almost as big as the set of all convex bodies in $X$.
So, your integral is $$\int_X (\inf\{t\in(0,\infty)\colon x/t\in C\})^2 \mu(dx)$$ for a general Minkowski set $C$.
Another way to write this integral is as follows: \begin{aligned}\int_X \|x\|^2 \mu(dx) &=\int_X \mu(dx)\int_0^\infty ds\,1\{s<\|x\|^2\} \\ &=\int_0^\infty ds\int_X \mu(dx)\,1\{s<\|x\|^2\} \\ &=\int_0^\infty ds\int_X \mu(dx)(1-\,1\{\|x\|^2\le s\}) \\ &=\int_0^\infty ds\,(1-\mu(\sqrt s B)). \end{aligned} We see that to compute your integral, you need to be able to compute the $\mu$-measure of an arbitrary Minkowski set, which is impossible to write explicitly even when $\mu$ is the standard normal distribution on $\mathbb R^d$.
