How much energy do you need to levitate/counteract gravity without any displacement or change in height?

A table can forever keep an apple "levitated" above the ground with it's normal force. That requires no energy. No work is done.

A force does not spend energy to fight against another force.

The force may cost energy to be produced, though. This is a separate issue. The jetpack spends fuel to produce an up-drift force, the human body spends nutrition to extend/contract muscles to produce the "holding"-force to hold a milk can, but the table spends nothing to produce it's normal force.

The jetpack falls down after a while and you feel tired after a while, not because work was done on the objects, but because work was done inside those "machines" (jetpack and body) that produce the forces. The table never gets tired. It never spends any work.

The issue is clearly not about holding anything. It takes no energy to hold stuff. You are correct that no work is done on the levitating man, if he undergoes no displacement. Work may be done inside the "machine" that produces the force, but that is internal.

For more, here's an article on the human body "machine" about this very topic.

The jetpack must create an upward force equal to that of gravity. For that reason, it expels some mass downward. Below, I calculate the work done by the jetpack assuming all mass expelled from it leaves at velocity $v$ which depends on the details of the jetpack's construction.

Suppose the jetpack of mass $m(t)$, where $t$ is the time loses $delta m(t)$ mass over a short time $delta t$, then its momentum changes by $delta p = delta m cdot v$, where $v(t)$ is the velocity by which matter is pushed down from the jetpack. To counterbalance the gravity, you have to have $m(t)g=-frac{delta m}{delta t} v$, where the sign comes from the fact that you lose mass from the jetpack as you try levitating. In the limit of very short time, the equation becomes:

$dfrac{dm(t)}{m(t)}=-dfrac{v}{g}dt$

This equation is solvable, and one can obtain the mass that needs to be lost to keep flying with the jetpack. Suppose you need to find the work done by flying it, that has to be equal to the kinetic energy of the expelled gas over some time. Since we assumed all mass is expelled at $v$ from the jetpack, the work should be $W(t)=dfrac{(m_0-m(t))v^2}{2}$, with $m_0$ being the initial mass of man+jetpack.

Your question is actually profound in a subtle way. The key to understanding this is that the man has a force being applied to him by gravity that is pulling him down. In order for him to stay aloft at a constant height, there must be a force that acts in the opposite direction and counteracts the force of gravity.

In your example, that counteracting force is supplied by the jetpack. So the jetpack must continually produce an acceleration upwards of equivalent to the weight of the man (and the jetpack.) But why is that different than when the man is standing on the ground? The earth's gravity is still acting on you but you don't have to continually burn fuel to stay in place. Consider the same standing on a large spring. When he first get's on it, he will move towards the earth and compress the spring until it pushes back enough to stop his motion. The spring's upward force is being supplied by weight of the man in a reflective manner. Essentially the ground does the same thing. The elasticity of the surface creates a mechanical equilibrium. Newtonian models don't really describe how materials produce elastic force. It is simply assumed.

To hover in the air conservation of momentum dictates that to keep a 100 kg object hovering we must "throw" 100 kg down towards the earth at a velocity of 9.8 m/s for every second we want to hover. The kinetic energy required to accelerate 100 kg to 9.8 m/s is 4.8 kilojoules. So a propeller that grabs 100 kg of air per second would need 4.8 kilojoules per second or 4.8 kilowatts (a watt is joules per second).

We could also propel twice the mass of air at half the speed to hover. Since kinetic energy is the square of the velocity propelling 200 kg down at 4.9 m/s would use 2.4 kilowatts or half the energy. So bigger is better and there is no theoretical limit to how low your energy consumption can go. Some sort of futurist tractor beam that can push or pull a very large mass of air would use almost no power. With our current technology and materials a very large open blade (i.e. a helicopter), large ducted fan or high-bypass turbofan are your best options as they will move the maximum amount of air with the minimum amount of energy.

If you want a more traditional jet pack where all of the reaction mass is kept on board then you're getting into rocketry and don't care about energy efficiency. You just care about the specific impulse (energy density) of the fuel. Also, even with the best rocket fuels your maximum flight times will be measured in seconds.

As @Steeven explains, no energy is required in principle. However, you will find that 'hovering' does require energy. How much depends on how you hover.

The basics are very simple. Gravity exerts a constant force $F$ on the levitating object. To counteract that force $F$, you can either place the object on a table, or impart momentum on an external reaction mass like air (helicopter), or propel some of your own mass (rocket).

The force generated by imparting momentum on an external reaction mass is

$$ F propto dot{m}v$$

with $dot{m}$ the reaction mass flow and $v$ the reaction mass velocity. To do this, you require a certain amount of power,

$$ P propto dot{m}v^2$$

From this, it is immediately obvious that you want to have a very large mass flow and a very low reaction mass velocity. This is why helicopters are more efficient than jetpacks (and turbofans more efficient than turbojets).

In rocket science, this still holds, but since you need to store all your mass $m$ on board before starting your hover, it is preferable to expend a lot of energy to minimise the mass flow. This is why jetpacks are still more preferable than rocket suits for hovering on Earth.