I'm confused on how they are doing this. It's a PNP BJT. According to
the LM3915 datasheet, the current this IC will be outputting at the
LED outputs is 12.5mA, since 12.5V/1k. Then 12.5mA going through the
1k resistors at the outputs give 12.5V again.
Yes, but you've fundamentally misunderstood the output stage of the LM3915. It does not source current, it sinks it. That is, the output pins look like an NPN transistor to ground. It is intended to work by effectively shorting the output pin to ground in order to draw current from a higher voltage. (Actually, it works as a current sink rather than a short, but this is a detail.)
To get an idea of this, the first thing you should have done was looked up the data sheet. Notice how the LEDs are connected in the Typical Applications (page 2).
With this said, this is actually a pretty crappy circuit. Each output pin, when active, will attempt to pull something like 15 mA from the bases of the PNP transistors. Since the LED current will be something like 55 mA (12 volts minus about 0.2 volts (PNP Vce) minus about 2 volts for the LEDs, divided by 180 ohms), a base drive of about 5 mA would be quite adequate. A good rule of thumb for driving BJTs as switches is to use a gain of 10. If the data sheet were followed with a base current of 5 mA calculated, each output would cause about 60 mW of power dissipation in the 3915 (12 volts times .005 amps), and the worst-case power dissipation in the chip would be about 600 mW, which would only occur when all segments are lit. For a normally-operating bar graph, this simply does not happen for any period of time.
So first, Vb = 12.5V. Ve = 12V from the supply. Veb should be .7V, but
this gives -.5V. Vb would have to be 11.8V to have a Veb of .7V
Nope, 12.5 is used in the data sheet as a maximum value. If you are actually using 12 volts, Vbe will be 0.7 (more or less) and Vb will be 11.3 volts. Keep in mind that the output stage of the 3915 will not produce a "higher" voltage - it always drops voltage.
Second it should be Ie = Ib+Ic, meaning the base current should be
flowing out, but the current is coming in from the IC.
Yes, Ie = Ib + Ic, but base current flows out from the base of a PNP. Base current will be a bit less than 12 mA, and LED (collector) current will in fact be about 55 mA, for a total emitter current of about 67 mA per lighted segment.
I tried making a simple overview of 1 LED, not sure if it's correct
Almost correct. The current source is pointing the wrong way. Furthermore, the 3915 output stage is not a perfect current source. so the base current will be less than 12.5 mA. In practice, the left-hand side of the base resistor will appear to be at a voltage of about a volt or so.
Third, let's say there was just 12V going to the LED's. The design the
person has uses 180ohms. (12-Vf)/180 = If. Red LED's I read generally
have around 2V Vf, so 12-2/180 is approximately 55mA. That seems
extremely high, and it also produces really high wattage.
Well-spotted. The 180 ohm resistor is a bit underpowered, as it is a 500 mW unit being asked to dissipate 550 mW. However, this is probably not an issue, since a 10% overpower condition is not usually fatal. Manufacturers do build making conservative assumptions about how the resistors will be used. If you put this in a small, closed box with little airflow, there may well be long-term problems, since the resistors will get hot. However, as I commented earlier, audio bar graphs spend most of their time with only one or two segments lit, so the heat will spread out some.
But, like I say, this is a pretty crappy version of the circuit, and should be a lesson to you about blindly grabbing circuits from the internet.