Olin Lanthrop clearly gave the most plausible explanation. But just for fun, let's just assume this was an electromagnetic trick. Would that be possible?

First - let's do this using electrical charge: how much charge would you need to allow levitation, and what would the potential have to be?

Some assumptions:

70 kg guy
40 cm levitation (based on apparent height)
equal charges
cane is perfect insulator

The force needed would be about 700N; solving

$$F=\frac{Q^2}{4\pi\epsilon_0 r^2}$$ gives $$Q = r\sqrt{4\pi\epsilon_0 F}$$ and $$V = \frac{Q}{4\pi\epsilon_0 r} \approx 3.5 MV$$

If I did not make a mistake, this makes the voltage on both the performer and the platform above which he is hovering about 3.5 MV. Note that this value is independent of the height he is hovering... just the force required. The breakdown voltage of atmospheric air depends on many factors - but you're not going to hold 3.5 MV on an irregularly shaped object without some serious corona discharge - the electrical breakdown strength of air is around $3.6\cdot10^6 V/m$, which will easily be exceeded in this configuration. I conclude it cannot be static electricity holding him up.

I will update with the magnetic calculation when I have a second...

Olin Lanthrop clearly gave the most plausible explanation. But just for fun, let's just assume this was an electromagnetic trick. Would that be possible?

First - let's do this using electrical charge: how much charge would you need to allow levitation, and what would the potential have to be?

Some assumptions:

70 kg guy
40 cm levitation (based on apparent height)
equal charges
cane is perfect insulator

The force needed would be about 700N; solving

$$F=\frac{Q^2}{4\pi\epsilon_0 r^2}$$ gives $$Q = r\sqrt{4\pi\epsilon_0 F}$$ and $$V = \frac{Q}{4\pi\epsilon_0 r} \approx 3.5 MV$$

If I did not make a mistake, this makes the voltage on both the performer and the platform above which he is hovering about 3.5 MV. Note that this value is independent of the height he is hovering... just the force required. The breakdown voltage of atmospheric air depends on many factors - but you're not going to hold 3.5 MV on an irregularly shaped object without some serious corona discharge - the electrical breakdown strength of air is around $3.6\cdot10^6 V/m$, which will easily be exceeded in this configuration. I conclude it cannot be static electricity holding him up.

So, could it be magnetism?

With magnetic levitation, there are two problems: the strength of the magnet, and stability (two dipoles cannot provide a stable levitation platform). Let's tackle the stability by making the platform is a superconductor; then the "magnetic pressure" is

$$P_{mag}=\frac{B^2}{2\mu_0}$$

Because I don't want to do the integral, I am going to assume that the field is uniform over a diameter equal to the distance; then we can compute the force:

$$F = P\cdot A = \frac{B^2 \pi d^2 / 4}{2\mu_0}\\ = \frac{B^2 \pi d^2 }{8\cdot 4\pi 10^{-7}}\\ = \frac{B^2 0.16 }{32\cdot 10^{-7}}$$

Setting to 700N and solving for $B$:

$$B = \sqrt{700 \cdot 200 \cdot 10^{-7}}\\ = 0.12 T$$

In principle it is possible to make permanent magnets that strong... but they would weigh quite a bit more than 70 kg (meaning you would have to update this calculation), and you wouldn't want to walk around with them on the market square of Leuven. Oh - and there's the minor problem of the cryogenics needed for your superconducting platform...