Is it possible to use forces on a current carrying conductor due to Earth's Magnetic field to levitate?

In principle it's possible, provided that the wire is held by some sort of flexible connections so that it isn't rigidly attached to the rest of the circuit. [If it were rigidly attached, R.W.Bird's answer would apply.]

Let us then consider a wire (not rigidly attached) of radius $r$, length $l$ and density $rho$. If the wire runs magnetic East-West, carries a current $I$ and the horizontal component of the Earth's magnetic flux density is $B_H$, then the vertical component of the magnetic force on the wire is $B_H I l$. So for levitation, $$B_H I l>mg text{that is} B_H I l>pi r^2 lrho g$$ So the current needed would be greater than $$I=frac{pi r^2 rho g}{B_H}$$ In the UK, $B_H=19 mu text T$ and $g=9.8 text{N kg}^{-1}$. Let's choose a copper wire, ($rho=9000 text{kg m}^{-3}$) of diameter 1.0 mm. Putting these figures into our formula gives $$I=3.6 text{kA}.$$ Just think what such a huge current would do to the wire in a very small fraction of a second! For this reason I'd say that levitation in the Earth's magnetic field isn't practicable; the field is just too weak.

Electric currents flow in loops (from a power supply and back). A loop of current forms a magnetic dipole. There is no net force on a dipole in a uniform magnetic field. The earth's magnetic field is uniform over a building sized volume.

Philip Wood gives a good introduction to the issues involved.

If instead of copper we use aluminium wire, and instead of diameter 1.0 mm we use a diameter 0.1 mm, then we get (see his answer for the formula) $I = 11$ amps which is feasible, except that this thin wire will not be able to sustain that current without heating up and melting. To calculate that we can use the resistivity $eta$ which leads to an electrical resistance $$ R = frac{eta l}{pi r^2} $$ and hence a dissipated power $$ I^2 R = pi r^2 l eta left( frac{ rho g }{B} right)^2. $$ Thus the dissipated power per unit volume of wire is $$ eta left( frac{ rho g }{B} right)^2. $$ For example, with our thin wire (diameter 0.1 mm) this gives 4 watts for each 1 cm length of wire. This will very quickly melt the wire unless we can get rid of this heat, and for that you need some sort of cooling method, such as water cooling, but the water cooling will itself be heavy so the whole thing will not lift off the ground.

So now let's consider superconducting wire. Superconducting wire cannot carry any amount of current, but it can carry very much higher currents than ordinary wire without dissipating heat. This kind of wire also needs cooling, for example by liquid nitrogen or liquid helium, but once cold it only heats up owing to ordinary heat conduction, so with good insulation it will not need so much added apparatus to keep it cold. I found in a quick investigation just now that for commercial wire you can get current densities around 20 kA/cm$^2$, and a tape of cross section 9.5 mm by $1.8$ mm can carry up to 1700 amps. Putting the numbers into the formula $$ I = ({rm area}) times rho g / B $$ I now get $I = 48000$ amps for this kind of wire. Unfortunately that is 28 times more current than the wire can carry. Another, thinner wire required a smaller current but again it was much more than the wire could carry without losing its superconducting properties.

However, the highest current densities that have been reported for superconducting materials are much higher, around 15 MA/cm$^2$. This suggests that in the near future superconducting tapes may become available that could carry enough current to levitate in Earth's magnetic field.

https://www.nature.com/articles/s41598-018-25499-1

Please note also the answer by R. W. Bird, which correctly points out that you can't make an independently flying device, like a flying carpet, this way. A horizontal coil, for example, will have a force directed up on one side and down on the other. You will have to have a wire trailing down to the ground to complete the circuit. The wire on the ground then gets pushed into the ground, while the wire in the air gets pushed upwards. Also, the device works better at the equator where the magnetic field is more horizontal. So I see this as maybe useful for moving stuff around a building site, or moving containers around a yard, or something like that, but not for transportation between cities.

Let us consider a thin circular aluminum wire in the Earth's equator plane at the altitude, say, 150 km, i.e., beyond atmosphere, so cooling is not an impossible problem. Aluminum is superconducting below 1.2 K. Let us assume that the superconducting current is I=0.3 A and the radius of the wire is r=10 micron. The magnetic field created by such current at the distance $r$ is $frac{mu_0 I}{2pi r}=0.006 T$, which is less than the critical magnetic field for aluminum (https://en.wikipedia.org/wiki/List_of_superconductors). The weight of $l=1 m$ of such wire ($rho=2700 kgcdot m^{-3}$) is $P=rhopi r^2 l g approx 8.3cdot 10^{-6}N$. The Earth's magnetic field on the equator is about $B=30cdot 10^{-6}T$, the force lifting 1 m of the wire in such field is $F=I l B=9cdot 10^{-6} N>P$, thus, levitation of a superconducting wire in the Earth's magnetic field is possible. Rotation of the wire around the Earth can add stability.

Why do I get results that are in tension with those of @Andrew Steane ? Because critical current of (type I) superconductors is proportional to radius, whereas its linear density is proportional to radius squared, so one can get better results for thin wires.