I went through your calculations and had to stop part way. I just wanted to know what is the heat input to each pad, in order to calculate the temperature difference when people are not using symmetrical pcb traces to the Kelvin pins. The ring is a second aspect to look at - I feel cuts/slices in the copper scuttle the intent to equalize heat outflow.
Missing is the heat loss due to convection currents within the plastic cap and radiation, verses conduction losses to any foam insulation around the IC.
Missing is the bonding wire(s) heat transfer die to pin.
Missing are glass seals' thermal resistance; dimensions unknown.
Schott has nice TO glass.
Just the Kovar pins:
Using 17.3W/m-K, 2mm long (shortest pins), 0.5mm dia pins=2mm^2.
Kovar Rt = L/(k*A) = (2mm/1000)/(17.3*(0.2mm^2/1000^2) = 578°C/W/pin. There's 8 of them for 72.3°C/W total.
Using 17.3W/m-K, 12.5mm long, 0.5mm dia pins=2mm^2.
Kovar Rt = L/(k*A) = (12.5mm/1000)/(17.3*(0.2mm^2/1000^2) = 3,613°C/W/pin. There's 8 of them for 451.6°C/W total.
Reality check, one of TiN's KX thermal cam pics showed a can temperature 46.93°C and PCB temp of around 34.53°C. The LTZ has long pins, say 12.5mm or 1/2".
LTZ1000 θJA=80°C/W value did not jive with the datasheet graphs
LTZ1000A θJA=400°C/W matches graphs OK though. θJA=175°C/W matches the (non-A) curves, LM723 is 165°C/W. That is reasonable for the measured temperature.
Using 0.13W (top of my head) to the die heater on an LTZ1000A, my pin calculations say the thermal resistance of the pins is too high for that temperature gradient so I must be wrong.