Compared to the usual acceleration of gravity that is almost 9.8m/s^2, it's only a 0.0048 = 0.48%. That is the slope of the real see compared to an ideal sea where there is no gravity from the ice. I'm using a spherical ice instead of a sheet of ice. With a sheet of ice the effect would be much smaller, but I'm too lazy to look up.
The gravity from the ice decrease as 1/r^2 when you go farther. We must integrate the slope to get the volume. I'm going to use a flat Earth, to simplify the calculation, but as the 1/r^2 reductions is quite fast, it's not a problem. With a curved Earth it will be more, but not too much.
Also, we must multiply by 2*pi*r to get the volume in the flat surface all around the sphere instead of the surface of a vertical cut of the sea.
The radios of the Earth is 6371km, so the "distance" to the north pole is 6371kmpi=20000km and we can cut the integral there https://www.wolframalpha.com/input?i=Integral+from+185+to+20... The result is 48,000 km^3 of water. But melting the Antarctic Ice sheet will release 24,300,000 km^3 So it's only a .2% more.
In other units, melting all the Antarctic Ice would increase the sea level like 58m (190ft). The additional effect of the gravity is less than .3m (1ft) And I expect that a calculation if a sheet shaped ice will give a much smaller result.
First, let's calculate the additional high. It's better to imagine the 0 is in the oposite pole, and the additional height is very similar to the old integral without the 2*pi*r https://www.wolframalpha.com/input?i=Integral+from+x+to+2000... i.e.
0.0048 * 185^2 * (1/r - 1/20000)
Now we must integrate it again, with the 2*pi*r Integral from 185 to 20000 of (0.0048 * 185^2 * (1/r-1/20000)), that is only 606km^3, so 1/80 of my previous calculation. So, fixing the last part of my comment with the new numbers:
But melting the Antarctic Ice sheet will release 24,300,000 km^3 So it's only a .0025% more. In other units, melting all the Antarctic Ice would increase the sea level like 58m (190ft). The additional effect of the gravity is less than .0004m (4mm, 1/8 inch)
The volume of the Antarctic Ice Sheet is 26.5 million cubic kilometres. If you use the Spherical Cow Theorem and put all that ice into an sphere, the radius is 185km https://www.wolframalpha.com/input?i=sphere+of+26.5+million+... . The gravity caused by that sphere at the surface is 0.047m/s^2 https://www.wolframalpha.com/input?i=gravity+of+24%2C380%2C0...
Compared to the usual acceleration of gravity that is almost 9.8m/s^2, it's only a 0.0048 = 0.48%. That is the slope of the real see compared to an ideal sea where there is no gravity from the ice. I'm using a spherical ice instead of a sheet of ice. With a sheet of ice the effect would be much smaller, but I'm too lazy to look up.
The gravity from the ice decrease as 1/r^2 when you go farther. We must integrate the slope to get the volume. I'm going to use a flat Earth, to simplify the calculation, but as the 1/r^2 reductions is quite fast, it's not a problem. With a curved Earth it will be more, but not too much.
Also, we must multiply by 2*pi*r to get the volume in the flat surface all around the sphere instead of the surface of a vertical cut of the sea.
The radios of the Earth is 6371km, so the "distance" to the north pole is 6371kmpi=20000km and we can cut the integral there https://www.wolframalpha.com/input?i=Integral+from+185+to+20... The result is 48,000 km^3 of water. But melting the Antarctic Ice sheet will release 24,300,000 km^3 So it's only a .2% more.
In other units, melting all the Antarctic Ice would increase the sea level like 58m (190ft). The additional effect of the gravity is less than .3m (1ft) And I expect that a calculation if a sheet shaped ice will give a much smaller result.