The existence of the exponential on $\mathbb{C}$ has a very basic, yet very strong consequence : $(\mathbb{C}^*,\cdot)$ is a quotient of $(\mathbb{C},+)$. This question is concerned with fields $K$ such that $K^*$ is a quotient of $K$ ; that is, with the existence of a surjective group morphism $K\to K^*$. I will refer to such morphisms as "exponentials" on $K$. (I know that the notion of exponential fields exists, but I only consider the surjective case.)
The existence of such a map is not benign : since the additive group $K$ is $q$-divisible for all $q\neq char(K)$, it implies that $K^*$ is also $q$-divisible. In characteristic $0$, this actually implies that $K$ is algebraically closed (I suspect that in characteristic $p$ this implies that $K$ is separably closed, but I'll focus on zero characteristic for now). EDIT : That was false, but it doesn't really matter since Eric's answer gives a construction when $K^*$ is divisible.
Question 1: It's a basic fact that algebraically closed fields of characteristic zero and with the same transcendance degree over $\mathbb{Q}$ are isomorphic. So $\mathbb{C}_p \simeq \mathbb{C}$ for all $p$, and thus $\mathbb{C}_p$ admits (at least one) exponential map. On the other hand, isomorphisms $\mathbb{C}\to \mathbb{C}_p$ are highly non-constructible objects, so this does not give us any clue about what an exponential on $\mathbb{C}_p$ may look like.
Can such an exponential map $\mathbb{C}_p\to \mathbb{C}_p^*$ be explicitly defined ? In particular, does it exist without the axiom of choice (or with only a weak version) ?
Question 2: If we put the axiom of choice back on :
For any countable cardinal $\kappa$, do algebraically closed fields of transcendance degree $\kappa$ over $\mathbb{Q}$ admit a surjective exponential ? (In particular, what about $\overline{\mathbb{Q}}$ ?)
We already know from $\mathbb{C}$ that this is true for $\kappa = 2^{\aleph_0}$ so since "algebraically closed fields of characteristic zero with a surjective exponential" may be countably axiomatized in first-order logic, Löwenheim-Skolem implies that there are models of all infinite cardinality, so in particular question 2 has a positive answer for all uncountable $\kappa$, and for some countable $\kappa$. But since all countable $\kappa$ give models of the same cardinality $\aleph_0$, we cannot use that remark to answer the question.