Capacitive coupling: how to collect 60uA from this set-up?
Back of envelope calculation coming, very low precision but the right ballpark.

1000 sparks/minute = 16 per second.

A current of 60uA = 60uC/second = about 4uC per spark.

To get 4uC from a 14kV pulse, we need a capacitance of about 4u/14k = 300pF.

That's a lot of capacitance to achieve by wrapping a bit of wire round your cable. In fact, it's equivalent to the capacitance of about 3m of 50ohm coaxial cable, which has broadly the dimensions you've specified in your OP.

So cover the 3m of the ignition lead with alli foil. Or replace the lead with 3m of 50 ohm coax, and take your coupled output from its screen.

Of course, an ignition waveform is not a nice sine wave, but has very steep edges. The current during the edge will be much, much higher than your target 60uA. But that's the capacitance you need if you want 60uA average. If you actually want 60uA peak, then a) you don't need as much capacitance to couple and b) you need to know a lot more about your ignition waveform, specifically the risetime into your load.

You might find that 300pF loads your ignition coil down somewhat. It might be better to extract your power from the coil's primary circuit, rather than the secondary.