You need to focus on how many times the instructions on the innermost loop will get executed. The outer loops are more like counters.

The inner-loop count will be as follows:

$i = 0; j = \emptyset$  
$i = 1; j = 0$  
$i = 2; j = 0, 1$  
$i = 3; j = 0, 1, 2$  
$\dots$

So you have a recurring sum of 1 for $j$ from $0$ to $i-1$, which can be mathematically expressed as:

$$\sum^n_{i = 1}\sum^{i-1}_{j = 0} 1 = \sum^n_{i = 1} i = \frac{n(n + 1)}{2}$$  
which is still $O(n^2)$.