Where does the energy from a parachute go? When a parachute slows the velocity of an object where does the energy go?
If it's a falling object the acceleration from gravity is roughly constant. How does air drag "dissipate" the extra energy?
 A: To add to HDE 226868's correct answer "heat":


*

*Even the ram drag component, arising when the parachute losslessly exchanges momentum with the relatively moving air and thus feels a Ram Pressure (see Wikipedia article of this name, and also my footnote), ends up as heat because the air eddies and currents airising from ram effect (see my footnote) then dissipate their energy by frictional drag with the surrounding air;

*A spacecraft re-entering the Earth's atmosphere gives a dramatic demonstration of this dissipation and this is why so much technology goes into heat shielding the spacecraft. The failure of this shielding lead to the destruction of the space shuttle Columbia in early 2003. Actually the main reason for the dramatic temperature rise here is a non dissipative (initially, at least) process like the ram pressure (see footnote).



Footnote on ram pressure: Ram drag is the (at first) non heating (non dissipative) drag exerted on a body as it losslessly exchanges linear momentum with the fluid it moves through. To understand the ram pressure, which arises particularly for supersonic objects, witness the object is just shoving fluid out of its way, and the latter flows off at some high angle to the trajectory. Think of a stationary object with a flat leading surface with a high speed flow around it. Fluid striking the flat surface gets deflected almost at right angles to the incoming flow. If you tally up the impulse per unit time that the object must be exerting on this flow to effect the change in fluid momentum, it is proportional to the flow rate (which, in turn, is proportional to the flow speed), and also proportional to the individual fluid particle momentum - also proportional to the flow speed. So the product of these two is proportional to $v^2$, where $v$ is the flow speed.
As noted by users GraphicsResearch and  user121330, during a spacecraft re-entry, another, related, initially non dissipative mechanism arises. The simple ram pressure description above describes the fluid as though it were incompressible, and is a good model in water. But air can be compressed and, during re-entry is so adiabatically as the ram effect unfolds. As the spacecraft does work on the air, the latter's temperature rises - enormously. The main mechanism here seems to be non-dissipative, adiabatic work done by the spacecraft in compressing the air. As long as the spacecraft can withstand this temperature of the cushion of air squashed before it, the latter actually shields the spacecraft ensures that all the frictional dissipation (which must ultimately dissipate all the energy) happens well away from the spacecraft: it is frictional, viscous loss in the air after the spacecraft has passed through. Thanks to the cushion there is almost NO frictional drag directly between the spacecraft itself and the air. The spacecraft sits on its cushion and the flow almost stagnates in the neighbourhood of the spacecraft's surface. This is why a blunt object feels much less heat load than a streamlined one during re-entry; see the "Blunt Body Entry Vehicles" section in the "Atmospheric Entry" Wikipedia page. A highly amusing discussion of this effect is also the discussion of whether a steak can be cooked by dropping it through the atmosphere at Article 28 "Steak Drop" at What-if.xkcd.com
A: Heat
Drag is the same thing as air resistance. It's a form of friction. Friction turns some of the kinetic energy of a moving object into heat; drag does the same thing, thus slowing a falling object down. When an object slows down due to friction, it heats up (and some of the heat dissipates to the surroundings). The same principle applies here: There will be some heating of the parachute and the air around it.
The drag equation governs the force of drag on an object. It is:
$$F_D=\frac{1}{2}\rho v^2C_DA$$
where $F_D$ is the force of drag, $\rho$ is the fluid's mass density, $v$ is the relative velocity of the object (relative to the fluid), $C_D$ is the drag coefficient, and $A$ is the area. In the absence of other forces, you can use the definition of work to figure out the work done by the drag:
$$W=Fs$$
$$W=\frac{1}{2}\rho v^2C_DAs$$
