This amplifier a clever way to inexpensively make a high quality amplifier, as it used relatively cheap tubes and no output coupling capacitor nor output transformer (nor even a power transformer for the output stage). So why not build it today, tarted up a bit with high quality parts and power supply regulation? Simply, I do not think it will work as well as it once did. Not in its original form that is.
Here is why.
The original amplifier used eight 12B4s and put out 12 watts into a 16 ohm load. Today we use 4-8 ohm speakers. The peak (non-positive grid) output current a triode is capable of delivering into a load resistance is given by
Ipk = B+ / (rp + Rload).
Since the rp of even four parallel 12B4s is much higher than any speaker impedance, the peak output current doesn't differ much between the 8 and 16 ohm load, but the peak output voltage is halved. In other words, the amplifier can only put out 5 watts into a 8 ohm load. Yes, 5 watts is more than any single 2A3 SE amplifier puts out and it is more than enough for a high efficiency midrange. (Now if the number of output tubes is doubled, the full 12 watts is possible. Since 12B4s are so cheap, doubling the number will not add much to the cost of the amplifier. Besides, as the amplifier runs in a lean Class-AB, each 12B4 dissipate less than 1 watt of heat at idle and should last forever.)
If you wish to have only a Class-A OTL, then more or different output tubes will be needed. Here is an example of seductive thinking: commercially made OTL amplifiers use eight 6AS7s and they supposedly put out 60 watts of Class-A power; thus, one 6AS7 can put out 8 watts of Class-A power. Seductive, as only the truly false and misleading can be. First of all, these amplifiers only put out about 4 watts of Class-A power. So using the ratio method, one 6AS7 should put out half a watt of Class-A power; it still can't. The math falls apart because of the squared terms in the power calculations: double the output current or the output voltage and you quadruple the output wattage.
Inversely, halving the output current or voltage, quarters the output wattage. This is the result of one variable's increase implying the increase of the other variable:
P = ½I²R
P = V² / 2R (RMS watts).