Exercise 1.2.11(b) on page 21 of Jiří Lebl's SCV notes defines a complex line through the origin as the image of a linear function $L:\mathbb C\to\mathbb C^n$, and asks the reader to:
Prove that if an entire holomorphic function is bounded on countably many complex lines through the origin then it is constant.
It seems obvious that for $n>2$, $\mathbb C^n$ contains infinitely many distinct complex lines through the origin (e.g. if $L_n:\mathbb C\to\mathbb C^2$ is defined by $L_n(z)=(z,nz)$, consider the collection $\{L_n(\mathbb C)\}_{n\in\mathbb N}$.) Perhaps I'm misunderstanding complex lines, however, because I believe I have a proof, using the Lebl exercise above, that $\mathbb C^n$ cannot contain infinitely many distinct complex lines through the origin for any $n$:
Suppose for that sake of contradiction that for some $n$ there is a countably infinite collection of linear functions $\{L_j\}_{j\in\mathbb N}$, $L_j:\mathbb C^n\to\mathbb C$, such that the complex lines $L_j(\mathbb C)$ are all distinct. Then each function $L_j(z)=(a_{j,1} z,\cdots,a_{j,n} z)$ can be extended to a function $\widehat L_j:\mathbb C\to\mathbb C^{n+1}$ by defining $\widehat L_j(z)=(a_{j,1} z,\cdots,a_{j,n} z,0)$. This yields a countably infinite collection of distinct lines in $\mathbb C^{n+1}$, on which the function $f(z_1,\cdots,z_n,z_{n+1})=z_{n+1}$ is zero. Thus $f$ is a holomorphic function $\mathbb C^{n+1}\to\mathbb C$ which is constant on a countably infinite collection of complex lines, hence constant by Exercise 1.2.11(b). This is, of course, absurd.
Everything you say is correct, and the exercise is incorrect. The exercise should be stated only in the case that $n=2$. Or, if $n$ is allowed to be greater than $2$, then the function should be required to be bounded on infinitely many hyperplanes through the origin (linear subspaces of dimension $n-1$), instead of on infinitely many lines.