$\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$Here is a computation of the Dobault cohomology of $X:=B(\infty) \setminus B(0) = \CC^2 \setminus \{ (0,0) \}$. I think that balls of finite radius should be behave basically the same way, but the details will be messier and it sounds like you just want to see an example.
Set
$$U_1 = (\CC \setminus \{ 0 \}) \times \CC,\ U_2 = \CC \times (\CC \setminus \{ 0 \}),\ U_{12} = (\CC \setminus \{ 0 \})^2.$$
So $\CC^2 \setminus \{ (0,0) \} = U_1 \cup U_2$ and $U_{12} = U_1 \cap U_2$. Each of $U_1$, $U_2$ and $U_{12}$ is a product of open sets in $\CC$, so they are Stein and thus any coherent sheaf on them has vanishing cohomology. We deduce by Leray's theorem that the Cech complex on $U_1$, $U_2$ computes Dolbeault cohomology of $\CC^2 \setminus \{ (0,0) \}$.
Let $\cO$ be the sheaf of holomorphic functions. So
$$\cO(U_1) = \left\{ \sum\nolimits_{j \geq 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}$$
$$\cO(U_2) = \left\{ \sum\nolimits_{i \geq 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}$$
$$\cO(U_{12}) = \left\{ \sum a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}.$$
We deduce that
$$H^0(X, \cO) = \left\{ \sum\nolimits_{i,j \geq 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}$$
$$H^1(X, \cO) = \left\{ \sum\nolimits_{i,j < 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}.$$
The sheaf $\Omega^1$ is a free $\cO$-module of rank two with basis $d z_1$, $d z_2$; the sheaf $\Omega^2$ is free of rank one with basis $d z_1 \wedge d z_2$. Thus,
$$H^0(X, \Omega^1) = \left\{ \sum\nolimits_{i,j \geq 0} a_{ij} z_1^i z_2^j d z_1 + \sum\nolimits_{i,j \geq 0} b_{ij} z_1^i z_2^j d z_2 \right\}$$
$$H^0(X, \Omega^2) = \left\{ \sum\nolimits_{i,j \geq 0} c_{ij} z_1^i z_2^j d z_1 \wedge d z_2 \right\}$$
$$H^1(X, \Omega^1) = \left\{ \sum\nolimits_{i,j < 0} a_{ij} z_1^i z_2^j d z_1 + \sum\nolimits_{i,j < 0} b_{ij} z_1^i z_2^j d z_2 \right\}$$
$$H^1(X, \Omega^2) = \left\{ \sum\nolimits_{i,j < 0} c_{ij} z_1^i z_2^j d z_1 \wedge d z_2 \right\}$$
where I have stopped writing down that the sums have to converge.
Let's see how this works with the Dolbeault-deRham spectral sequence. The $(p,q)$ term is $H^q(\Omega^p)$. The maps on the first page go $(p,q) \to (p+1,q)$ and are induced by $d$. So we need to compute the cohomology of
$$H^0(X, \cO) \to H^0(X, \Omega^1) \to H^0(X, \Omega^2)$$
$$H^1(X, \cO) \to H^1(X, \Omega^1) \to H^1(X, \Omega^2).$$
Put a $\mathbb{Z}^2$-grading on these vector space where $z_1$ and $dz_1$ are in degree $(1,0)$ and $z_2$ and $dz_2$ are in degree $(0,1)$. Ignoring the convergence conditions for a moment, these complexes are direct sums of finite dimensional complexes in each degree $(i,j)$, and we compute that they are exact except when $(i,j) = (0,0)$. The complexes in degree $(0,0)$ is $\CC (1) \to 0 \to 0$ and $0 \to 0 \to \CC (z_1^{-1} z_2^{-1} dz_1 \wedge dz_2)$.
I am going to omit the argument for length, but the convergence conditions don't change anything: The next page of the spectral sequence is $\CC(1)$ in position $(0,0)$ and $\CC (z_1^{-1} z_2^{-1} d z_1 \wedge d z_2)$ in $(2,1)$. There can be no arrows from $(0,0)$ to $(2,1)$ on future pages, so the complex collapses, and we compute that $H^0(X) \cong \CC$ and $H^3(X) \cong \CC$. This is of course correct, since $X$ retracts onto the $3$-sphere of radius $1$.
