Calculating current in a series circuit
Let's look at Ohms Law \$ V  = IR \$ 

What this means is that the voltage between two points, is the current that is between those 2 points, and the resistance between those 2 points. 

In your analysis you did the following:
$$ I = \frac{V}{R} = \frac{3.3}{220} = 15mA $$

Since voltage is a potential difference, the 3.3V is with respect to ground. So the two points we are looking at, is the supply rail (3.3V) and ground (0V). The resistance between these two points as initially placed, is 220 ohms. That means that the LED has NO resistance at all. And this is where it broke down. You made an assumption that the LED had no resistance, and it does (it's a weird type of resistance called dynamic resistance - but don't worry about that one right now).

But when an LED has current passing through it, it has with it a forward voltage drop that will approximately be constant. Depending on the color of the LED, it will have different forward voltage drop. Red led's are about ~2V.

So if we try again with this new knowledge, we can come up with the following:

$$ I = \frac{V}{R} $$
$$ I = \frac{\Delta V}{R} $$
$$ I = \frac{3.3V - 2V}{220} $$
$$ I = \frac{1.3V}{220} = 5.9mA $$