[RyanN]

Any reason not to use a resistor connected to the 5V Analog Breakout?

Let's do the math.

V=IR
R=V/I

Voltage drop needed across the resistor will be 5V - 3.3V = 1.7V

From the data sheet, it uses 500μA typical. I'm not sure how the sleep current will work out.

So..
R = V/I
R = 1.7V/500μA
R = 1.7/500x10^-6
R = 1.7/0.0005
R = 3400Ω = 3.4kΩ

Now, the accelerometer has a bit of flexibility. You should be good with any resistor between:

V = 5V-2.2V = 2.8V
R = 2.8V/500μA = 5.6kΩ

and

V= 5V-3.6V = 1.4V
R = 1.4V=500μA = 2.8kΩ

I would stay away from either extreme, preferably lower since I'm not sure what will happen with the sleep current.

Before doing this though, let some of the more experience engineers take a look at what I provided before going with it. You could get a 3.3V voltage regulator, and it will work fine. Resistors seemed simpler and easier to acquire (@ RadioShack or in your shop).