Let $D$ be a strictly convex domain in $\mathbb C^n$ i.e $T_p(\partial D)\cap \bar D=\{p\}$. The real dimension of $T_p(\partial D)$ is $2n-1$ and thus it contains a complex linear subspace $S_p(\partial D):=T_p(\partial D)\cap iT_p(\partial D)$ of complex dimension equals to $n-1$. Since $S_p(\partial D)$ is complex subspace of $\mathbb C^n$ of codimension $1$ it follows that there exists a complex linear map $l:\mathbb C^n\rightarrow \mathbb C$ such that $S_p(\partial D)=l^{-1}(0)$.
Now consider the function $f(z):=\dfrac {1}{l(z)}$ which is holomorphic on $\mathbb C^n\setminus l^{-1}(0)=\mathbb C^n\setminus S_p(\partial D)$. But $\mathbb C^n\setminus S_p(\partial D)\cong \mathbb C$ and $S_p(\partial D)\cap \partial D=\{p\}$.
My question is: How to show that $f$ can't be extended holomorphically to $p\in \partial D$?
After translating and rotating, you may as well assume $p = 0$ and $\ell(z) = z_{n}$ (i.e., $S_{p}(\partial D) = \mathbf{C}^{n-1} \times \{0\}$), so that $f(z) = \frac{1}{z_{n}}$. Since $f$ is unbounded in every neighborhood of the origin $0$, $f$ does not extend holomorphically to $0$.