Let $R$ be the ring $\mathbb{Z}_p[\zeta]$, where $\zeta = \zeta_{p^k}$. Let $\pi$ be a uniformizer of $R$, so that $|\pi|^{(p-1)p^{k-1}} = |p|$.

Write $D_r$ for the disk $\{x \in R: |x - 1| < |\pi|^r\}$. Then the morphism $\log$ is an isomorphism on $D_{p^{k - 1}}$, which has index $p^{p^{k-1}}$ in $D_0$.

On the other hand, the roots of unity in $D_0$ are exactly all the $p^k$-th roots of unity. Consequently, in the case $k = 1$, one concludes that $D_0$ is isomorphic to $\mathbb{Z}/p\mathbb{Z} \times D_1$.

Let $R$ be the ring $\mathbb{Z}_p[\zeta]$, where $\zeta = \zeta_{p^k}$. Let $\pi$ be the uniformizer $\zeta - 1$ of $R$, so that $|\pi|^{(p-1)p^{k-1}} = |p|$.

Write $D_r$ for the disk $\{x \in R: |x - 1| < |\pi|^r\}$. Then the morphism $\log$ is an isomorphism on $D_{p^{k - 1}}$, which has index $p^{p^{k-1}}$ in $D_0$.

On the other hand, the roots of unity in $D_0$ are exactly all the $p^k$-th roots of unity. Consequently, in the case $k = 1$, one concludes that $D_0$ is isomorphic to $\mathbb{Z}/p\mathbb{Z} \times D_1$, and the image of the $\log$ map is the disk $\{x \in R: |x| < |\pi|\}$.

To get a feeling of what happens in the general case, let us look at the case $p = 3$ and $k = 2$. The group $\mu$ of roots of unity in $D_0$ has $9$ elements, but the quotient $D_0 / D_3$ has $27$ elements. So the quotient $D_0 / \mu D_3$ has three elements, generated by $1 - \pi$. One calculates: \begin{eqnarray*} \log(1 - \pi) & = & \log(1 - \pi)(1 + \pi) \\ & = & \frac{1}{3}\log(1-\pi^2)^3 \\ & = & \frac{1}{3}\log(1 - \pi^6 + \cdots) \\ & = & \frac{1}{3}(-\pi^6 + \cdots) \end{eqnarray*} which has absolute value $1$.

So in this case, the image of $\log$ consists of three disks, contained in $\mathfrak{m}$, $1 + \mathfrak{m}$, $2 + \mathfrak{m}$, respectively, and all of the same size as $D_3$.

The general case should be similar. For given $p$ and $k$, the image can be calculated with an algorithm. It then depends on what other information one needs.

In the case $R = \mathbb{Z}_p[\zeta]$ where $\zeta = \zeta_{p^k}$:

Let $\pi \in R$ be a uniformizer (e.g. $\pi = \zeta - 1$), so that $|\pi| = |p|^{1 / (p-1)p^{k-1}}$. Then every element in $1 + \mathfrak{m}$ has a $\pi$-adic expansion, of the form $1 + a_1 \pi + a_2 \pi^2 + \cdots$, with $a_i \in \{0, 1, \cdots, p-1\}$.

It is then easy (!) to see that the subgroup $D_r := \{x : |x - 1| < |\pi|^r\}$ has index $p^r$ in $1 + \mathfrak{m}(= D_0)$.

For $r = k$, the logarithm converges on the disk $D_k$, which has index $p^k$ in $D_0$. On the other hand, the group of roots of unity in $D_0$ is exactly all the $p^k$-th roots of unity.

Thus in this case we have a decomposition: $D_0 \simeq \mathbb{Z}/p^k \mathbb{Z} \oplus D_k$. Hence the image of the logarithm is exactly the disk $\{x: |x|<|\pi|^k\}$.

Let $R$ be the ring $\mathbb{Z}_p[\zeta]$, where $\zeta = \zeta_{p^k}$. Let $\pi$ be a uniformizer of $R$, so that $|\pi|^{(p-1)p^{k-1}} = |p|$.

Write $D_r$ for the disk $\{x \in R: |x - 1| < |\pi|^r\}$. Then the morphism $\log$ is an isomorphism on $D_{p^{k - 1}}$, which has index $p^{p^{k-1}}$ in $D_0$. 

On the other hand, the roots of unity in $D_0$ are exactly all the $p^k$-th roots of unity. Consequently, in the case $k = 1$, one concludes that $D_0$ is isomorphic to $\mathbb{Z}/p\mathbb{Z} \times D_1$.