Voltage regulator

Yes, the dropout voltage is the headroom the regulator needs to work with. This is the minimum amount the input voltage must be above the output voltage for the regulator to work properly. So as you say, a 3.3 V regulator with 1.1 V dropout requires at least 4.4 V input.

Stuff happens, so it's usually a good idea to give the regulator a little more room. However, going too far will be less efficient, so there is a tradeoff. The range you quote of 4.75 to 5.25 volts could well be a reasonable tradeoff. Basically, that's a "5 V" supply.

The output current of a linear regulator is its input current (minus the small current on the ground pin, but that is small enough to ignore for the point I'm trying to make). Since the input voltage is higher than the output but the currents are the same, the output has less power. The difference in input power and output power is wasted in the regulator as heat. Another way to look at this is that the regulator will dissipate the difference between the input and output voltages times the current as heat. For example, consider a case where the input is 5.0 V, the output 3.3 V, and the current 500 mA. The voltage accross the regulator is 5.0V - 3.3V = 1.7V. That times 500 mA is 850 mW, which will go into heating the regulator. That would be fine for something like a TO-220 package, but a SOT-23, for example, would burn up at 850 mW.

Firstly, your use of a 7812 voltage regulator for you initial circuit may have seemed fine when you tested it but the 7812 needs at least 14V (or above) for it to provide a regulated output of 12V - if you check what the output is when the input drops significantly below 14V you will see that I'm correct. At 11.58V, it's likely that your "regulated" output is less than 11V.

I'm saying this because you have a misconception as to how linear voltage regulators work and you are in danger of believing you can apply this to higher voltage supplies like the 48V in your question.

You have a desired input voltage range of 43.2V to 67.2V and your required output is 48V. Without compromising your intended requirements, the only way this can be achieved is with a buck-boost switching regulator.

In addition to this, powering fans in series will likely destroy one fan - it will go open circuit (or burning) and then nothing will work. To simplify this, consider using a switching buck regulator that converts your 43.2-67.2V to +12V, then power your fans in parallel.

Maybe try one of these: -

I'm assuming you don't need more than 1A at 12V output as per your original design.

EDIT A higher output version is this and now, you will have to use external transistors on the device: -

I would also seriously consider the LTC3810. Here's a picture and also see figure 19 in the data sheet: -