What would it take to pressurize a cabin with surface pressure?
Not much. The current requirement for 8,000 ft equivalent pressure translates into 8.11 psi at 43,000 ft. Boeing studied the benefits of lower pressure and found that going below 6,000 ft equivalent pressure brought no additional benefits. 

The full pressure differential between sea level and 43,000 ft is 12.3 psi or just 52% more than that at 8,000 ft. This would require a heavier fuselage, but would not be impossible to do at all. Statistical methods give a mass factor for the fuselage pressure differential of $(1 + ∆p)^{0.203}$ which yields 1.0933 for 8.11 psi (∆p = 0.552 atm) and 1.1314 for 12.3 psi (∆p = 0.837 atm). Increasing the pressure to its sea level value would increase the mass of a typical airliner fuselage by 3.5%.

If we use the volume of the cabin, we can also calculate how much heavier the air is: The A330-200 has a cabin volume of 372 m³, and with the density of air at 8,000 ft of 0.9633 kg/m³, this computes to 358.35 kg. Compared to the mass of the same volume at sea level (455.7 kg), the difference is roughly equivalent to the mass of a single passenger (not counting the amenities carried along for her/him, too). In other words: The added mass of air for sea level pressure will make no difference if one passenger more misses her/his flight that day.