Based on the Philips data sheet about the ECC88, this tube can only be used with a heating voltage of 6.3 volts (between pin 4 and pin 5). Pin 9 is the shield between the two triodes and not the center tap of the two heaters connected in series. It is therefore not possible to operate the heating filaments of the two triodes separately.
The heater current is 365 mA (not 300 mA).
The eBay device you mentioned can supposedly deliver 15 mA at 280 VDC and 1500 mA at 6.3 VDC.
The total output power is therefore:
PPlate voltage = U*I = 280*0.015 = 4.2 W.
PFilament voltage = U*I = 6.3*1.5 = 9.45 W.
Let's assume that the entire device causes a heat loss of 20%.
The supplying power supply must therefore provide:
(4.2 W + 9.45 W =) 13.65 W. This corresponds to (100 - 20) = 80% of the power consumption. 100% is therefore ((100*13.65)/80 =) 17 W.
The supplying power supply must therefore provide a current of I = P/E = 17/12 = 1.4 A at an output voltage of 12 VDC. As others mentioned, the heater starting current is more than the nominal value; I would at least double the output power of the power supply, i.e. specify a power supply with an output of 12 VDC / 2.8A.
Your 12 VDC / 1A power supply is definitely too weak and needs to be replaced with one with around 2.8A or more.
e.g. Meanwell GSM36B12-P1J
https://uk.farnell.com/mean-well/gs...ical-ac-dc-12v-3a/dp/2815798?ost=gsm36b12-p1j
or
https://tinyurl.com/5av8sh6j
Nick Salis