Look for a solution of the form $f(z)=e^{\lambda z}$. Plugging this to your equation, you obtain that $\lambda$ must be a zero of the entire function $$F(\lambda)=\sum_{j=1}^n e^{\lambda\eta_j}.$$ When there are at least two distinct $\eta_j$, this entire function $F$ has infinitely many zeros $\lambda_k$. Then any finite linear combination $$f(z)=\sum_{k} c_ke^{\lambda_k z}$$ is a solution. Taking sufficiently small $c_k$, you can pass to the limit and obtain also infinite linear combinations. If $F$ has a multiple root $\lambda_k$ there will be also solutions of the form $P_k(z)e^{\lambda_k z}$, where $P$ is a polynomial.

There is a theorem of Malgrange (Theorem 16.4.1 in Hormander's Analysis of linear partial differential operators) which says that these are all solutions. (Linear combinations of exponentials are dense in the set of all solutions).

Since the coefficients $c_j$ can be arbitrary (subject to the condition that the series converges), you can obtain entire solutions of arbitrarily fast growth.

Look for a solution of the form $f(z)=e^{\lambda z}$. Plugging this to your equation, you obtain that $\lambda$ must be a zero of the entire function $$F(\lambda)=\sum_{j=1}^n e^{\lambda\eta_j}.$$ When there are at least two distinct $\eta_j$, this entire function $F$ has infinitely many zeros $\lambda_k$. Then any finite linear combination $$f(z)=\sum_{k} c_ke^{\lambda_k z}$$ is a solution. Also limits of such functions will be solutions. If $F$ has a multiple root $\lambda_k$ there will be also solutions of the form $P_k(z)e^{\lambda_k z}$, where $P$ is a polynomial.

There is a theorem of Malgrange (Theorem 16.4.1 in Hormander's Analysis of linear partial differential operators) which says that these are all solutions. (Linear combinations of exponentials are dense in the set of all solutions).

Since the coefficients $c_j$ can be arbitrary (subject to the condition that the series converges), you can obtain entire solutions of arbitrarily fast growth.