Wattage to produce 6L of air from Water in less than a second?
I guess it would have to be the amount energy released by the reaction $$\ce{2H2 + O2 -&gt; 2H2O}$$ where you have four liters of hydrogen gas and two liters of oxygen gas, all divided by one second (or less, increasing the wattage as you decrease the time). 

You asked about "air" — but you can't get normal air from electrolysis of water. 

So, the heat of combustion for hydrogen is $286\ \mathrm{kJ/mol}$. $$286\ \frac{\mathrm{kJ}}{\mathrm{mol}} \times \frac{4\ \mathrm L}{22.4\ \frac{\mathrm L}{\mathrm{mol}}} = 51.1\ \mathrm{kJ}$$ and that is assuming atmospheric pressure which is certainly low, but how low depends on your depth below the surface of the water. So with the "barely under the surface" energy of $51.1\ \mathrm{kJ}$, we can then divide that by seconds: $$\frac{51.1\ \mathrm{kJ}}{1\ \mathrm s} = 51.1\ \mathrm{kW}$$ so you would need a bare minimum of $51.1\ \mathrm{kW}$ of power to convert water into a 67:33 mixture of $\ce{H2:O2}$. That seems like a lot of energy in a small amount of time to me: think about a six-liter balloon with the perfect combustion ratio of hydrogen and oxygen. That reaction is a lot faster than one second, but it's still a lot of energy. Tough to get into a small, portable battery and have it fill a capacitor with that amount of e energy every few seconds.

I'll be honest, I'm just doing these calculations on-the-fly, and I may have made a mistake or five, but it seems reasonable to me. If anyone takes any issue with my answer, please, feel free to edit/give suggestions.

Edit: This Wikipedia page has some information on hydrogen-oxygen diving air. They use a 96:4 ratio. Not sure why, but if that's the best ratio, this would take significantly more energy per breath, since a lot of oxygen would just get dumped.