Is escape velocity the same for all objects? Would a lighter-than-air craft in the mid atmosphere at 80,000 feet altitude need to achieve the same velocity to escape earth gravity as the space shuttle?
 A: Yes. Escape velocity does not depend on the flying object's mass, but only on that of the earth. The precise formula is given by 
$$v_e=\sqrt{\frac{2GM}{R}},$$
where $G$ is the gravitational constant, $M$ the mass of the earth and $R$ the object's distance from the center of the latter. 
This is of course only true if one ignores the interaction between the atmosphere and the vehicle. 
A: Assuming no air resistance, friction, etc., the answer is usually. @FredericBrunner is pretty much correct in his analysis, if we are assuming the object being talked about is the only object in the equations. However, if there are 2 or more objects cooperating or competing in getting away from Earth's gravitational pull, things get slightly more complicated. As noted in answers to this question (similar but not exactly a duplicate), if one object is going at a very low speed, but another object is pushing it along with a force sufficient to accelerate both objects to escape velocity and keep it there, both objects will escape Earth's gravitational pull. Therefore, there is technically no such thing as "escape velocity"; it's more like "escape momentum" or "escape force" (or work, or any other combination of speed & mass).
The main reason either object would not be near escape velocity but would still escape the Earth's pull, however, is usually because the forces pushing the object upward are not much greater in total than the sum of the forces pushing the object down. In absolute terms, if the object is of mass m and the acceleration required to maintain escape velocity assuming no external forces is a, the difference between the forces causing the object to move upward and the forces causing the object to move downward would be between 0 and ma. Thus we normally don't see this very much.
To answer your question, any two objects at the same distance from Earth need the same velocity to escape the Earth's pull if the net forces acting on the respective objects are identical in magnitude and direction. Technically, they need the same velocity to escape at each moment, as once the difference between the forces gets to be less than the force required to escape gravity, they of course start moving towards the Earth again.
I haven't gone into everything involved, but I hope this was a good start. As I said, the previous answer was close enough in most circumstances, and of course my answer also fails in some situations.
A: Given that the spacecraft of mass $m$ is moving initially with escape speed $v_e$ away from Earth, and given that the speed of Earth (of mass $M$) is initially $0$, 
therefore the center of mass of these two objects moves always at speed $$u = \frac{m}{M + m} v_e$$
(with respect to the system in which Earth had initially speed $0$).
In an idealized "escape experiment" both the spacecraft and Earth should eventually be moving almost at (and ever closer to) the same speed $u$ as the center-of-mass system; that's when the (non-relativistically approximated) initial kinetic energies of the spacecraft and of Earth (with respect to the center-of-mass system) are almost (and ever more fully) converted to (also approximate) gravitative potential energy.
Equating these initial kinetic energies and gravitative potential energy:
$$\begin{eqnarray}\frac{m}{2} (v_e - u)^2 + \frac{M}{2} u^2 & = & \frac{G~M~m}{R} \\ \frac{m}{2} \left(\frac{M}{M + m} v_e\right)^2 + \frac{M}{2} \left(\frac{m}{M + m} v_e\right)^2 & = & \frac{G~M~m}{R} \\ \frac{v_e^2}{2}~m~M~\left( \frac{M}{(M + m)^2} + \frac{m}{(M + m)^2}\right) & = & \frac{G~M~m}{R} \\ \frac{v_e^2}{2}~m~M~\left( \frac{1}{M + m}\right) & = & \frac{G~M~m}{R} \end{eqnarray},$$
and consequently
$$v_e = \sqrt{ \frac{2~G~(M + m)}{R} }.$$
In practical cases the mass of the spacecraft may of course be negligible compared to the mass of Earth; so
$$v_e \simeq \sqrt{ \frac{2~G~M}{R} }.$$
A: exactly as @user12262 said, the mass does count as a factor, but it's so much smaller than the mass of the earth, that it's negligible. the other answers just point at the formula, but that's not the entire formula (Again, as @user12262 stated). The entire formula does have the second object's mass, but it's not needed, as the earth has so much more mass.
think back to 7th grade. remember that experiment your science teacher showed you? the one that stated a bowling ball and a feather will fall at the same speed if its in a Vacuum? Well, that's not entirely true. you see, the equation for gravitational force is:
Fgrav= G (m1 * m2 / r2)
or
force of gravity = the gravitational constant times mass1 times mass2 divided by the distance squared
with the variables:
G = 6.674E-11
Mass of Earth = 5.972E24 kg
Average Mass of Bowling Ball = about 2kg
Average Mass of Feather = about .05g or 5.0E-5kg
distance from center of earth at sea level = 6378km
so the force of gravity on a bowling ball from sea level is:
6.674E-11 * 5.972E24kg * 2kg / 6378km2 = 19.5962N
the force for the feather is:
6.674E-11 * 5.972E24kg * 5.0E-5kg / 6378km2 = 0.000489905N
As you can see there is a significant difference in forces
