Electronic – Considering power loss in a simple series connection due to cable

For a more general system I would do this differently. Maybe it's overkill in your case, but this isn't that hard. I've done something similar, although the power requirements were higher.

In my system I used CAN as a multi-drop bus to all the nodes. In the common configuration it uses differential signalling, which is a good idea for long distances where you could pick up common mode noise. I reserved one pair for the CAN lines. The other three pair were used for power, with a power and ground on each pair. That way the total common mode current of each twisted pair will still be close to zero. In my case I used 48 V because that is the maximum where you generally don't have to worry about safety issues much. There are plenty of microcontrollers with CAN built in, and the silicon deals with collisions and retries automatically.

In your case the power requirements are less, so 24 V might be a good choice for the power. Lots of transistors and buck regulator chips work up to 30 V. 28 V would be better if you want to push the most power and still use the cheapest power supplies at the ends, but I said 24 V because you're not pushing the limit and that's a very commonly available off the shelf power supply voltage.

At each node, put a small buck regulator. These are small and cheap nowadays. The MCP16301 can handle up to 30 V in, costs under $1, includes the switch, and comes in a nice and small SOT-23 package. Make sure to put a decent ceramic cap right at the input to the buck regulator though. The 24 V needs to be low impedance at high frequencies for the buck switcher to work. You probably want something like a 10 µF 30 V cap.

The advantage of this scheme is that you can tolerate a fairly wide power voltage at each node, but due to the higher voltage and therefore lower current, you won't have much drop. There will also be less heat at each node since the buck switcher will be more efficient than a linear regulator after you raise the power voltage enough to cover all the worst case conditions with headroom.

Another significant advantage is that there will be less ground offset between nodes, again due to the lower power current at the higher voltage.

Your power estimates are way too high like you say.You should measure or google or get the inductance of the coil .Once you have L then in your formula it goes .Its now an LR circuit where the current builds slowly starting from zero and limiting out given dwell time at your DCR value .With sensible parameters the coil finishing current wont get that high so your average coil current which is your current consumption on inductive ignition system will be much lower than your estimate .Realy your estimate assumes that L is zero or can be neglected which is not true here .