There is a more general question here of which this one is a special case:
There are three answers there giving different answers, but I argue that my answer is the correct one :) Namely, if the covariance matrix of the data is $\newcommand{\S}{\boldsymbol \Sigma} \S$ and if we consider a unit vector $\newcommand{\w}{\mathbf w} \w$, then the variance explained by the projection on this vector is given by $$R^2=\frac{\|\S \w\|^2}{\w^\top \S \w \cdot \mathrm{tr}(\S)}.$$
This question asks about a single variable (e.g. the first one), which means that $$\w = (\begin{array}{}1&0&...&0\end{array})^\top.$$
Plugging it in the general formula, we obtain that $$R^2 = \frac{\sum \sigma_{1k}^4/\sigma_{11}^2}{\mathrm{tr}(\S)},$$ where $\sigma_{ij}^2$ are the elements of $\S$.
Note that if the first variable is uncorrelated with all the others (as is the case for PCA eigenvectors), i.e. $\forall \sigma_{1k}^2=0$ for $k\ne 1$, then the formula reduces to the well-known PCA expression: $$R^2 = \frac{\sigma_{11}^2}{\mathrm{tr}(\S)}.$$
Alternative derivation
We can obtain the same result via a different route. The proportion of variance of the $k$-th variable explained by the first variable is given by the square of the correlation coefficient $$R_{12}^2 = \rho_{12}^2 = \frac{\sigma_{12}^4}{\sigma^2_{11}\sigma^2_{kk}}.$$ The amount (not the proportion) of explained variance is given by $R_{12}^2\sigma_{kk}^2$. Taking the sum over all variables and dividing by the total variance, we obtain the same expression as above: $$R^2 = \frac{\sum R_{12}^2\sigma_{kk}^2}{\sum \sigma_{kk}^2} = \frac{\sum \sigma^4_{1k}/\sigma^2_{11}}{\mathrm{tr}(\S)}.$$
