The other answers give fine examples, but I found a reference to the 1981 thesis of Livné which constructs complex hyperbolic lattice which admits a surjective holomorphic map to a Riemann surface. This is detailed in chapter 16 of

*Deligne, Pierre; Mostow, George Daniel*, **Commensurabilities among lattices in $\text{PU}(1,n)$**, Annals of Mathematics Studies. 132. Princeton, NJ: Princeton University Press. 183 p. $ 49.95/ hbk; $ 19.95/pbk; £ 33.50/hbk; £ 15.00/pbk (1993). ZBL0826.22011.

Since Riemann surfaces have covers with arbitrarily large betti numbers, so do the corresponding ball quotients. The point is here that if one has a holomorphic map $\phi: X \to Y$, $X$ and $Y$ compact, $Y$ a Riemann surface, then $\pi_1(X)$ must surject a finite-index subgroup of $Y$. Otherwise $\phi_\#(\pi_1(X))$ would induce an infinite cover $\tilde{Y}\to Y$ and lift $\tilde{\phi}: X\to \tilde{Y}$ so that $\phi$ factors through $\tilde{\phi}$. But the map $\tilde{\phi}$ must be constant, since it maps a compact complex manifold to a non-compact Riemann surface (by the open mapping theorem, restricted to 1-dimensional complex subspaces of $X$, $\tilde{\phi}$ is open if non-constant, and hence $\tilde{\phi}(X)$ is both open and compact in $\tilde{Y}$, a contradiction). Thus, $\pi_1(X)$ must surject $\pi_1(\tilde{Y})$ for $\tilde{Y}\to Y$ a finite-sheeted cover. Then covers of $\tilde{Y}$ with arbitrarily large betti number induce such covers of $X$.

Some generalizations to other examples are given by Deraux using a forgetful map.

If the lattice is arithmetic, then covers induced from a map to a Riemann surface will usually not be congruence covers.

In the arithmetic case, once one has a (congruence) cover with positive betti number, one can find further (congruence) covers with arbitrarily large betti numbers, as hinted at in Venkataramana's answer. Another perspective on this is given here.