You're misunderstanding what you need to drive a transistor.
A transistor will allow up to a specific amount of current through the collector depending on the current that flows through the base. If less is needed by the load then less will flow.
Think of it like a tap. If there's not much water pressure it doesn't matter how far you turn on the tap, after a certain point the flow of water won't increase any more.
To allow 60mA through the collector you need to provide at least 0.6mA (assuming a hFE of 100) through the base. That 0.6mA will set the upper limit of the current to 60mA. If you provide more than 0.6mA then you will allow a higher upper limit.
The 0.6V on the base is the voltage (difference between base and emitter when wired as "common emitter") at which current will start flowing through the base. It's the same concept as the forward voltage of an LED. For switching the voltage you put on the base must be more than 0.6V to allow current to start flowing. Since 5V is more than 0.6V you don't need to worry about it. If you want precision in your calculations you will want to take that 0.6V into account, though.
So if you want at least 0.6mA from 5V with a 0.6V drop on the BE junction, you need:
R=V/I = (5 - 0.6) / 0.0006 = 7.3kΩ.
That's the maximum resistance you want. Anything more than that and the current through the base will limit the current through the collector to below the 60mA you want. So you must select some resistance that is lower than that.
What value? Well, it really doesn't matter that much. The lower the value the higher the current, but at these levels the current levels you're dealing with will be minuscule anyway.
It's common to use a 1kΩ resistor in the base for a transistor when using it as a simple switch like you are. That would give you, for each transistor:
I=V/R = (5 - 0.6) / 1000 = 4.4mA
For all 8 outputs that's a paltry 35.2mA. Far below the total the chip can provide of 70mA.
If you care about current consumption (you're running from batteries, for instance) you may want to increase to 4.7kΩ resistors to reduce the total drive current down to 7.5mA.