Is is OK to drive AD620 with +-5V and expecting the output range up to 3.3V?
Summary:


What Oli said, plus
You cannot just use a 5V and ground power supply.
Worst case range is 1.4V below +ve supply and 2.1V above negative supply.
To JUST handle 3.3V output you need at least +4.7V and -2.1V supply. 
Use of a switched capacitor power supply will provide the negative rail with ease.




What Oli said.
Plus:

If you want to use +5V wrt ground and no other supply then   

Vout+ = 5 - 1.2V = 3.8V.
 Vout - = 0V +1.1v = 1.1V  

SO your Vout swing with 5V single supply = 1.1V to 3.8V.
 Note that this is for typical unspecified condions (see table in Oli's answer) and in worst case conditions the results are worse. 

From the datasheet (bottom of page 3) the worst case over temperature is about from 2.1V above the most negative rail to 1.4v below the most positive rail. So to JUST get 0-3.3V operation you need supplies of
V+= 3.3 + 1.4 = +4.7V
V- = 0 - 2.1V = - 2.1V.  

Supplies of say +5V and -3V would probably suffice.
A -3V supply van be easily obtained from a single stage "diode" pump = switched capacitor = charge pump power supply operating from +5V input. eg 

The LT1054 - datasheet here is more capable than some more basic switched capacitor  converters- providing eg voltage regulation, but can be used in basic mode as well if desired. The diagrams below show (top diagram) basic mode with the IC plus one capacitor providing a negative supply from 5V in, and a regulated mode in the lower diagram.





Addition -  Output range and device performance 

I accidentally jumped data sheet tables in my to and from checking (I know this as I recall the S/C current in the table when I started) and as Steven notes - I used the figures from the table on page 3 which is for inputs.   BUT I started on page 4 / outputs and the conclusions I drew are the same but figures vary slightly.

IF you are trying to use an ADC on an input that swings from 0 Volt to + 3.3V then the AD620 needs to be able to cover this range.

I'm not sure that Steven understood which way the AD620 range limits lay relative to supply - he probably did - BUT the  result of using +/- 2.3V supplies is to ensure that the IA cannot cover the range required. BUT with +/- 5V supplies it can. I read the original query as a +5/0 V supply - I assume I missed the +/- but it may have been edited since. SO

With a +- 5V supply all is sweetness and light.  


Top of page 4 ( :-) ) says output can swing from 1.1 V above lower rail to 1.2 V below upper rail at +/- 5V at unspecified temperature = -3.9V to +3.8V at +/- 5V supply
Below that the full temperature spec says 1.6V above V- to 1.5V below V+ = -3.4V to + 3.5V (just !!!) at +/- 5V supply.
To JUST meet 0-3.3V spec at full temperature you can use a -1.6V to +4.8V supply. Obviously you'd aim at more than that.


With an instrumentation amp you can examine input specs and output specs separately. IF the input meets all input specs then with appropriate gain and biasing you can make the output swing over the whole legal output range. 

Datasheet dependability:

The reason that 


"But I still have wonder that when the output is near +Vs-1.2V, will it lose linearity? How serious is that? I could not find info from the datasheet..."


is not of concern is that, as long as all input and output parameters are in spec the IA will meet thelinearity, distortion gain etc specs that the manufacturer specifies. If the IA lost performance in someway as Vout or Vin or whatever approached a limit then the datasheeet would (should) say so. if it makes no mention then you can assume that performance is as specified. Note that this assumes that the datasheet is truthful and contains no errors. For a reputable manufacturer (such as Analog Devices) the first assumption is a reasonably safe one and the second is usally safe and you have to assume it is error free until proven otherwise. Some manufacturers data sheets or products are not so reliable or error free.