The existence of analytical branch of the logarithm of a holomorphic function

I am happy to report that, yes, you can define $\log f$ even in much greater generality than in your situation.
Namely, if you have a real differentiable manifold $M$ and a function $f\in \mathcal C^\infty (M)$ never zero on $M$, then $f$ has a $\mathcal C^\infty $ logarithm as soon as the first De Rham cohomology group of $M$ vanishes: $H^1_{DR}(M, \mathbb R)=0$.

The definition of the logarithm is straightforward: fix a point $x_0\in M$ and define $$(\log f)(x)= \int_\gamma \frac {df}{f} $$
where $\gamma$ is a differentiable path joining $x_0$ to $x$, along which we can integrate the closed $1$-form $\frac {dg}{g}$.
The vanishing cohomology hypothesis ensures that the value of $\log f$ at $x$ does not depend of the path $\gamma$ chosen.

If $M$ happens to be a complex holomorphic manifold and if $f\in \mathcal O(M)$ is holomorphic, then the logarithm of $f$ will automatically be holomorphic: $\log f\in \mathcal O(M)$.
This applies to your case since a simply connected manifold -and a fortiori a convex set in a vector space- has zero first De Rham cohomology group.

Finally, just for old times' sake, let me sum this up in the language of classical physics :
$$\text{Every conservative vector field has a potential}$$

A variant
Specialists in complex manifolds are addicted to the exact sequence of sheaves on the complex manifold $X$: $$0\to 2i\pi \mathbb Z\to \mathcal O_X \stackrel {exp}{\to} \mathcal O^\ast_X \to 0 $$

A portion of the associated cohomology long exact sequence is $$ \Gamma(X,\mathcal O_X) \stackrel {exp}{\to} \Gamma(X,\mathcal O^\ast_X) \to H^1(X,\mathbb Z) $$
which shows again that every nowhere vanishing holomorphic function on $X$ is an exponential (in other words: has a logarithm) as soon as $H^1(X,\mathbb Z)=0$.

Edit: I misread the question and only address the case $n=1$.

If you allow the function $f$ to have roots, then the answer is clearly no. There is no analytic branch of the logarithm of $z$ in any neighborhood of $0$.

If you exclude roots, the answer is yes.

More generally:

Let $f: G \to \mathbb{C}$ be holomorphic with $0 \notin f(G)$ and assume that $G$ is a simply connected domain (in particular $G$ convex suffices). Then $f$ has a holomorphic logarithm $g: G \to \mathbb{C}$, that is $f(z) = e^{g(z)}$ for all $z \in G$.

Fix $z_0 \in G$ arbitrarily, choose $g(z_0)$ in such a way that $f(z_0) = \exp{g(z_0)}$ and put $g(z) = g(z_0) + \int_{z_0}^{z} \frac{f'(w)}{f(w)}\,dw$. Check that $g$ is well-defined, holomorphic, and that $\frac{d}{dz} [f \cdot e^{-g}] = 0$, hence $f = e^g$.