The info about 100/150mA USB capability in PC is incorrect. Any normal (not "low-power embedded") USB host must supply 500mA (for USB2.0 ports) and 900mA for USB3.0 ports. In a typical PC this power comes from standby +5VSB rail from system Power Supply. Power rating for this rail is no less than 2A for AT12 compliant PC. Of course it is not advisable to take all this power from one port, due to contact rating limits, and possible polyswitch overcurrent protection of USB ports.
Therefore, if the attached USB device draws total less than 1-1.5A, the design should work fine even if PC is in standby mode (turned off but still connected to AC outlet).
However it is still preferable to use a stand-alone power for this kind of projects. The Arduino board must be designed in a way that, if connected to external higher-voltage source, it would not backdrive any significant current into USB VBUS pin. Usually it gets accomplished by means of one single Schottky diode between VBUS and internal power rail on Arduino board.
You can check if your Arduino drives anything back: power the Arduino with, say, 12V power supply, and check with a Multimeter if VBUS pin on the (unattached) USB connector has any voltage. You should load the VBUS with 10k-15k resistor, and the voltage should be small, 200mV or less, if I remember correctly.
ADDENDUM1: Just checked with Arduino-nano schematics. Yes, there is a "auto selecting" Schottky diode, so there will be no backvoltage on VBUS. However, the Arduino design violates another USB requirement - no signal on any connector pin when VBUS is inactive (low). In the Arduino-nano the VBUS does not control the state of FT232 bridge, so it will always pull-up D+ (USBDP) line as soon as 5V is applied, which is in violation of specs. The side effect of this back-driving is that your PC might not re-connect with the Arduino after waking-up, although it is a rare case these days.