The answer is $\frac{1}{4}$, which is attainable by letting $a_i = \frac{1}{2}$ for all $i$. Now it suffices to show that there exists $i$ such that $a_i - a_i a_{i+1} = a_i(1 - a_{i+1}) \leq \frac{1}{4}$. Simply take the least $a_i$ and notice that $a_i(1 - a_{i+1}) \leq a_i(1 - a_i) \leq \frac{(a_i + (1 - a_i))^2}{4} = \frac{1}{4}$.

If any of $a_i$ is zero or one, than obviously the minimum is zero. So assume $a_i \neq 0$ and $a_i \neq 1$ for all $i$. Note that $\frac{1}{4}$ is achievable with $a_i = \frac{1}{2}$. Now we show that it cant be more than that. We have $(a_1 - a_1a_2)(a_2 - a_2a_3) \dots (a_n - a_na_1) = a_1(1 - a_1) \dots a_n(1 - a_n) \leq \frac{1}{4^n}$ by AM-GM. And since there are $n$ numbers, their minimum must be less than or equal to $\frac{1}{4}$, as desired.

Assume that $a_{1}=a_{n+1}$. We know that $a_{i}\in [0,1]$ , so $a_{i}-a_{i}a_{i+1}=a_{i}(1-a_{i+1})\geq 0$, so we can look at the $n$-th root of the product of the $n$ cyclic expressions and notice that: $$\min\{a_{1}-a_{1}a_{2}, a_{2}-a_{2}a_{3},\dots ,a_{n}-a_{n}a_{1}\}\leq \sqrt[n]{\prod\limits_{i=1}^{n} a_{i}-a_{i}a_{i+1}}=\sqrt[n]{\prod\limits_{i=1}^{n} a_{i}(1-a_{i})}\overset{AM-GM}{\leq} \sqrt[n]{\prod\limits_{i=1}^{n} \frac{1}{4}}=\frac{1}{4} $$This value is achieved when all of the AM-GM inequalities reach equalities, so $a_{i}=\frac{1}{2}$ $\forall i=1,2,\dots ,n$ and this indeed gives an example for $\min\{a_{1}-a_{1}a_{2}, a_{2}-a_{2}a_{3},\dots ,a_{n}-a_{n}a_{1}\}=\frac{1}{4}$ as $a_{i}-a_{i}a_{i+1}=\frac{1}{4}$ for all $i$.