How to find the operating current consumption of the op-amp?
The power an op-amp dissipates is from two parts.

1, power to run the device. The 8mA value listed is the current the device will consume with the output open circuit and applies between the rail voltages. So in your case, power will be the voltage range 10V times the quiescent current, here 8mA, So it will dissipate 80mW just sitting there.

2, Power to run the load. 

This gets a bit more complicated. 

If you are driving a load to ground at say 10mA, then that current will either be sourced from the +5v rail or soaked up by the -5V rail, so you need to add 10mA * 5V, another 50mW, in this example. 

If the load you are driving goes to the bottom or top rail, then the worst case would be 10mA times your rail range, 10V. so you would need to add in 100mW. 

Of course the latter is worst case with a given load.

If you are using the op-amp to buffer a DC value this worst case value may apply. If you are using the op-amp to drive some AC signal you can use the RMS value of the signal in the above math. 

However, be aware, when you undersize the device based on some estimated average signal condition, if there is a failure or wrong connection that changes that signal, your device can cook. If there is ever any doubt on what the signal will be, use the worst case numbers.

When it comes to power dissipation, you really want to derate the devices specified power rating by a good margin, perhaps even less than 50% of that. Other-wise the device will be very hot and you will need to add heatsinking.

Further, you really need to factor in the tolerances of your design. For example, if your power supplies are rated at 5% your 10V becomes 10.5V. You need to use the worst case numbers here.