How much open circuit voltage can a C.T. generate if burden is removed
I dug an odd toroidal transformer core out of my junkbox to do some quick sums. This one has an area 13x32 mm = 384mm2, and a mean diameter of 60mm, so 0.188m circumference. You could probably wind a transformer in the 80-100VA range on it. I have wound current transformers on similar cores.

If we use a peak flux Bmax of 1.8T at 50Hz, the max rate of change of flux is
\$2\pi f B_{max} \$ = 565T/s, which for an area of 384mm2 gives 217mV/turn peak, call it your round 200mV!

What will a single turn of current carrying wire drive the field to, if there is no secondary current flowing, i.e. no burden resistor?

Pass a single turn through the core, and pass 1A through it. As the core has a circumference of 0.188m, that's an H field of 5.3A/m. That will create a flux in the core of B = \$\mu\$H = \$4\pi 10^{-7}\mu_rH\$ = \$6.7\mu T\$ per \$\mu_r\$ per amp.

What to use for the permeability of iron? 2k is a minimum for typical transformer irons, but current transformers tend to be wound with high \$\mu\$ types to improve the accuracy, so perhaps 5k is a more reasonable figure. That means a flux of 0.033T per amp.

To generate 200mV/turn, we need a peak B field of 1.8T, which needs 1.8/0.033 = 54 amps peak, or 38 amps rms. 

My shower takes 40 amps rms. If I put the kettle, toaster and the oven on at the same time the water heater is on, that comes to more than 40 amps. I only have an 80A company fuse, so I do think before putting the kettle on if there's someone in the shower!

1kV with 5000 turns and 'domestic' current levels, no problem

What happens if more primary current flows? I initially thought 'the core saturates, so limits the voltage'. But no. The core saturates, which flattens the sinusoidal core flux waveform into a more trapeziodal shape, increasing the rate of change through the zero crossings, which increases the output voltage.