I've never seen actual figures but, in general, articles I've seen about flight state that "most" lift is generated from the angle of attack and relatively little from the Bernoulli effect. I suspect the exact figures are rather variable and probably depend on whether the plane is climbing, descending, banking, etc and will also vary from plane to plane. Maybe this is why exact figures seem not to be quoted.
The pressure difference between the top and bottom of the wing is quite real, though note that on the top of the wing it's not a vacuum as the pressure doesn't decrease that much. The lowered pressure above the wing will indeed tend to pull the skin off the wing, or more precisely the air within the wing that is at normal atmospheric pressure will try to push the skin off. Once again I can't give you exact figures - I must admit I thought ballpark figures would be easy to calculate, but Google has failed me.
Incidentally, there's a good NASA article on this subject at http://www.grc.nasa.gov/WWW/k-12/airplane/wrong1.html and it even includes a Java applet for you to play with the details of the wing. A longer slightly more staid article is at http://www.free-online-private-pilot-ground-school.com/aerodynamics.html
Later:
If an approximate answer would be OK then you could could use Bernoulli's equation as described in http://en.wikipedia.org/wiki/Bernoulli%27s_equation#Incompressible_flow_equation. Although this really only applies to incompressible fluids, and air is obviously compressible, the article suggests it would be a reasonable approximation for low speeds.
Rewriting the equation to make it more useful for our purposes gives:
$$P = \rho A - \rho \frac {v^2}{2} - gh$$
where $A$ is some constant and $h$ is the height. We don't know the constant, but let $P_{bot}$ be the pressure below the wing and $P_{top}$ be the pressure above the wing then we can take the difference between them i.e. the pressure drop between the bottom and top of the wing. If we assume the height is constant i.e. we can ignore the thickness of the wing we get:
$$\Delta P = P_{bot} - P_{top} = 0.5 \rho (v_{top}^2 - v_{bot}^2)$$
I don't know what speed you plane flies at, but let's guess at 30 m/s and let's guess that there's a 10 m/s difference between the air speed at the top and bottom of the wing, so that's $v_{bot} = 30$ and $v_{top}$ = 40. Google gives the density of air at ground level as 1.225 kg/m3.
$$\Delta P = 0.5 \times 1.225 \times (40^2 - 30^2) = 429 Pa$$
429 Pa is 4.29 grams per square cm or 0.06 pounds per square inch, so it's completely insignificant.
You probably know the equation for the drag, but just for record it's:
$$ F_{drag} = \frac{1}{2} \rho \space C_d A \space v^2 $$
and rearranging this gives:
$$ \frac{2F_{drag}}{\rho \space C_d \space v^2} = A $$
and you're given $C_d$ (1.4) and $v$ (3 m/s). I would guess you're meant to take the density of Earth's atmosphere at sea level, and a quick Google gives this as 1.2754 kg/m$^3$ (is the density of the Martian atmosphere really 2/3 that of Earth? I thought it was more like 1%). The force is just the weight of the probe (40kg) multiplied by the acceleration due to gravity at the surface of Mars ($g_{Mars} = 3.75\ m/s^2$). So you have everything you need to calculate $A$.
It's not obvious to me why you need the speed of the probe without the parachute ...
Best Answer
Physics should not be different on other planets, so the same laws apply as on earth. Only the results of an optimization might look unfamiliar. See here for an answer on Aviation SE on a Mars solar aircraft.
The lift slope equation you found is only valid for slender bodies, like fuselages and fuel tanks, and once wing span becomes a sizable fraction of length, more complicated equations will be needed, and Mach number effects must be considered, too. See here for a more elaborate answer.
Generally, to fly like on earth would mean that the ratio of dynamic pressure and mass is the same. Then you would use the same aircraft as on earth (provided the other planet's atmosphere contains enough, but not too much oxygen for the engine to function).
Dynamic pressure $q$ is the product of the square of airspeed $v$ and air density $\rho$: $$q = \frac{\rho\cdot v^2}{2}$$
Lift is dynamic pressure times wing area $S$ and lift coefficient $c_L$ and must be equal to weight, that is the product of mass $m$ and the local gravitational acceleration $g$: $$L = q\cdot S\cdot c_L = m\cdot g$$
The lift coefficient is a measure how much lift can be created by a given wing area and can reach values of up to 3 in case of a landing airliner. Then the wing uses all kinds of high-lift devices (slats, slotted flaps), and once those are put away, the lift coefficient of an airliner is at about 0.5. For observation aircraft, less speed is required, and a normal lift coefficient for them would be 1.2. I see no reason why this number should be different just because the atmosphere is different.
The most important number would be the Reynolds number $Re$. It is the ratio of inertial to viscous forces in a flow and is affected by the dimensions of your plane (on earth we use the wing chord $l$) and the density and dynamic viscosity $\mu$ of your planet's atmosphere. $$Re = \frac{v\cdot\rho\cdot l}{\mu}$$
Lower Reynolds numbers will translate into higher friction drag, which depresses the maximum achievable lift-to-drag ratio. Gliders fly at Reynolds numbers between 1,000,000 and 3,000,000 and airliners can easily achieve 50,000,000. When you need to optimize for a more gooey atmosphere, your wings will become less slender than on earth, because you will enlarge wing chord $l$ to keep $Re$ up.
Once you need speed to get the weight lifted, the Mach number $Ma$ might become important. Generally, subsonic flight is the most efficient, and it has a natural limit at $Ma^2 \cdot c_L = 0.4$. This is what can be achieved with todays technology. The speed of sound in a gas is mainly a function of temperature - Mach 1 on Mars is 238 m/s.
The first part of an airplane to hit a Mach limit are the propeller tips. Maybe you need to have several slow-spinning, small propellers than one big, honking propeller which would provide the best efficiency as long as its tips are well below Mach 1.
Last, you need to know the number of atoms per gas molecule. Air is dominated by two-atomic molecules, but maybe your planet has an atmosphere like early earth with lots of carbon dioxide. This will affect the ratio of specific heats $\kappa$ - this means the rate of heating and cooling with compression and expansion of the gas might be different than on earth. This will come into play when you approach or exceed Mach 1.