Big Oh Time Complexity Involving 'for i in range(n)'
So you have the right logic, if you have a loop of $O(n)$ iterations, the complexity of the entire loop will be $O(n*\text{complexity of loop operations})$. In this case, you again are correct that your loop's complexity is $O(n)$ for each iteration. 

Your last bullet point shows that you understand this as well, as the total loop complexity is then $O(n^2)$. However, this trivially results in an overall time complexity of $O(n^2)$, not $O(n)$ as you concluded.

I imagine what you were really asking about is the complexity of range(n). While it is true that the initial call to range is $O(1)$, that is because it produces a generator which takes $O(1)$ to return the next successive value i.e. it doesn't produce the entire range right away, but instead on demand. However, keep in mind that the generator will be advanced $O(n)$ times (once for every iteration of the loop).