For each $k = 2, 3, \dots, n-1$, define $S_k = \sum_{i=1}^{n} \angle A_i A_{i+1} A_{i+1}$ where the indices are taken modulo $n$. Note that
$$S_2 + S_3 + \dots + S_{n-1} = \angle A_1 + \angle A_2 + \dots + \angle A_n = (n-2) \cdot 180^\circ \quad (1)$$
because each angle $\angle A_i A_j A_k$ with distinct $i, j, k$ is counted exactly once. By definition, the given sum $S$ is at least $S_k$ for each $k = 2, 3, \dots, n-1$. Therefore, by equation (1), we obtain
$$S \ge \frac{S_2 + S_3 + \dots + S_{n-1}}{n-2} = 180^\circ.$$
On the other hand, if $A_1 A_2 \dots A_n$ is cyclic (for example, it is a regular $n$-gon), then
$$f(1) + f(2) + \dots + f(n) = A_1 A_2 + A_2 A_3 + \dots + A_{n-1} A_n = 180^\circ.$$
Therefore, the smallest possible value of $S$ is $180^\circ$.

(In the figure below, we have $n=5$. Also, $S_2$ is the sum of all angles in white, $S_3$ is the sum of all angles in grey, and $S_4$ is the sum of all angles in black.)