Is every holomorphic function in a neighborhood of the origin of $C^n$ necessarily bounded? Is every holomorphic function in a neighborhood of the origin of $C^n$ necessarily bounded?
In which case the statement can be true?
Thanks in advance
 A: The answer is no. If the neighbourhood $U$ is unbounded, at least one of the coordinate functions is a counterexample.
If the neighbourhood is bounded, let $p\in\partial U$ be a point of maximal distance $r$ from the origin. Let $l(z) = a_1z_1 + \cdots + a_nz_n + b$ be an affine function such that $l(p)=0$, but $l(z)\neq 0$ on $U$. (Choose the coefficients so that the zero set of $l$ is contained in the tangent space of $B_r(0)$.) Finally, let $$f(z) = \frac1{l(z)}.$$
This construction shows that at least one point of $\partial U$ is pseudoconvex.
A: Per the request of the OP and the clarification of the original question (see the comment thread above), here's my answer:
Question: Let $U$ be a neighborhood of the the origin in $\mathbb{C}^n$. Under what conditions on $U$ is every holomorphic function on $U$ bounded?
Answer: No such conditions exist. Let $U$ be a neighborhood of the origin in $\mathbb{C}^n$ and let's assume $U\neq \mathbb{C}^n$ (if $U=\mathbb{C}^n$, then take identity function). The boundary of $U$ is $\text{Bd}(U)=\overline{U}\setminus U$ and is non-empty. Let $a\in \text{Bd}(U)$ and choose the function $\frac{1}{z-a}$. It is holomorphic on $U$ but unbounded.
Question: If $f$ is a holomorphic function on a neighborhood $U$ of the origin, then does there exist an open subset $V\subseteq U$ such that $f$ is bounded on $V$?
Answer: Yes, always. Chose a neighborhood $V$ of the origin with compact closure such that $\overline{V}\subseteq U$. Holomorphic functions are continuous and continuous functions on compact sets are bounded. So, the restriction of any holomorphic function on $U$ to $V$ is bounded.
I hope this helps!
