As long as the mass that we aren't dropping is very large and is kept constant, then the mass of the object we are dropping has no considerable effect on its acceleration. This is because of Newton's 2nd Law:
$$F = ma$$
Where $m$ is the mass that is accelerating, i.e. the smaller mass we are dropping. So, if $F = G\frac{Mm}{r^2}$, where $m$ is the mass we dropped, and $M$ is the big mass that the object we dropped is fall to, then:
$$a = \frac{F}{m} = G\frac{M}{r^2}$$
So, while acceleration is dependent in $M$, it does not depend on the mass of the dropped object.
The constant value $g$ is actually only true on the earth's surface, and is appropriately defined as:
$$g_{earth} = G\frac{M}{(R_{earth})^2}$$
Where $R_{earth}$ is the radius of the Earth.
Notice that I said the bigger mass, $M$ (or, the mass that is causing the gravitational field) is, indeed, big. If it were not that big, the object of the mass we dropped (by Newton's 3rd Law) would cause a force on $M$ that results in a significant acceleration of $M$. This means that both masses are significantly accelerating towards each other, so that the effective acceleration of the dropped object (i.e. the acceleration in a reference frame where $M$ is stationary) would be the sum of the magnitudes of the accelerations of each object in a non-accelerating frame of reference:
$$a_{effective} = a_m + a_M = G\frac{M}{r^2} + G\frac{m}{r^2} = G\frac{M+m}{r^2}$$
In which case, the mass of the dropped object does indeed affect the effective acceleration.
(Note that the "effective acceleration" is the acceleration of the dropped object when $M$ appears stationary to the observer. This does not appear to obey Newton's 2nd Law (i.e. $a_{effective} \neq a_m$ because this frame of reference is accelerating, and accelerating reference, otherwise know as non-inertial reference frames, do not obey Newton's 2nd Law! In fact, this question assumes that, in the inertial frame of reference, the two masses are heading towards each other in a straight line...)
