Digit LED has single input, multiple outputs for each segment. Concerns about "sinking" 5v supply into digital pins
The device you have is a common anode display. This means that the anodes (+ve terminals) of all the LEDs are connected to one common pin (well, 2 pins actually, pin 3 and 8 are connected internally) . Their cathodes (-ve terminals) are what make up the remaining pins. Now, to light up any LED, you need 5 V at its anode and GND at its cathode. You have connected 5 V to pin 3, which stands for the common anode of every LED, so all that is left for any LED to turn on, is for you to connect GND to its corresponding cathode. So when you made all the remaining pins (the cathodes) OUTPUT, by default the Nano sets those pins to a logic LOW (or GND) value, so ALL the LEDs got lit up. Each LED now has 5 - 0 = 5 V across its terminals and is forward-biased. 

However, when you write HIGH to any cathode, the potential difference across the associated LED/segment is now zero, because that LED has 5 V at its anode (pin 3/8) and 5 V at its cathode (as a result of writing HIGH), and 5 - 5 = 0 V so it is reverse-biased (at least 2 V or thereabouts is needed for a red LED to become forward-biased) and so the segment turns off. And this is how you control which segment is on at any time: You write a LOW to a cathode to turn on the associated segment, and write HIGH to turn it off. It sounds counter-intuitive but it follows if you think about it. Regarding your question, there's no damage being done to anything; the minimum voltage across the LED is 0 V (well below its maximum peak inverse voltage), while the maximum voltage across the LED is around 2 V, if you use the highly recommended current-limiting resistors (220/330 ohms is good) in series between each cathode and its Arduino digital pin.

Without any resistors, you will be drawing a lot of current through each pin, in total, a lot more than the Nano can safely supply. Use resistors that limit the current through each LED to 10-20mA.