How much longer is the path through spacetime of a mass that falls freely compared to a resting mass? A mass that falls to Earth follows a shortest path through spacetime. If a mass falls from a 1km high building, how much longer will its path be compared to a mass resting on a table?
 A: It only makes sense to compare path lengths for worldlines that have the same start and end points in spacetime. The worldlines of the object on the building and the object on the table have different starting points, so you can't usefully compare their lengths.
If you say "let's assume the worldlines start at the same time, even though they are in different locations in space" then you have to define what "at the same time" means. This will depend on the motion of the observer, so the comparison of worldline lengths that have different starting points is observer-dependent.
A: Richard Feynmann furnished a tangible example of how to estimate this sort of thing. Imagine the circumference of the earth, as measured within its gravitational potential energy "well". If you left your tape measure in position but caused the earth to suddenly vanish so you were no longer in that well, the length of the tape measure would change by about 1/4".
A: 
[...] If a mass falls from a 1 km high building, how much longer will its path be compared to a mass [being held] on a table?

A direct and relevant comparison would be of

*

*the duration $\tau P[ \, \text{launch from }T, \text{impact on }T \, ]$, or for short: $\tau P_{TT}$, of a projectile $P$, from being launched from the table (on the surface of the Earth), flying freely (without air friction etc.) straight up to 1 km high above the table, and continuing to drop freely straight back to impact on the table, and


*the duration $\tau T[ \, \text{launch of }P, \text{impact of }P \, ]$ of table $T$, from having launched the projectile until being impacted by the projectile; for short: $\tau T_{PP}$.
Considering the (nearly) spherical geometry of Earth's surface, and the geometry of spacetime above being (nearly) uniform on the scale of few km, with constituents of Earth's surface, of buildings, etc. being held rigid wrt. each other and supported with acceleration $g_T \approx 9.81~{\rm m/s^2}$ against "falling down", the equivalence principle allows to estimate the ratio $\frac{\tau T_{PP}}{\tau P_{TT}}$ for an equivalent configuration in a flat region:
participant $Q$ being and remaining at rest, as a member of an inertial system in the flat region, and participant $H$ being constantly accelerated (a.k.a. moving hyperbolically) with $a_H = g_T$, such that $H$ and $Q$ pass each other twice.
Then we have explicitly:
$$\frac{\tau T_{PP}}{\tau P_{TT}} \approx \frac{\tau H_{QQ}}{\tau Q_{HH}} = \frac{2 \, (c / a_H) \, \text{ArcSinh}[ \, (a_H / c) \, (\tau Q_{HH} / 2) \, ]}{\tau Q_{HH}} \approx \qquad \qquad \qquad \qquad \qquad \qquad \\ \qquad \qquad \qquad \qquad 1 - \frac{1}{24} \, \left( (a_H / c) \, \tau Q_{HH} \right)^2 + \frac{3}{640} \, \left( (a_H / c) \, \tau Q_{HH} \right)^4 - \dots$$
where (of course) $(a_H / c) \, \tau Q_{HH} = (g_T / c) \, \tau Q_{HH} \approx (g_T / c) \, \tau P_{TT} \ll 1.$
Further, the distance between $Q$ and member $A$ of the same inertial system (a.k.a. at rest wrt. $Q$) who is just reached and passed by $H$ and momentarily co-moving with $H$ as $H$ reached its apex (wrt. the inertial system to which $Q$ qnd $A$ belong together) is explicitly
$$QA = (c^2 / a_H) \, \left( \sqrt{ 1 + \left( (a_H / c) \, (\tau Q_{HH} / 2) \right)^2 } - 1 \right) \approx \qquad \qquad \qquad \qquad \qquad \qquad \\ \qquad \qquad \qquad \qquad (c^2 / a_H) \, \left( \frac{1}{8} \, \left( (a_H / c) \, \tau Q_{HH} \right)^2 - \frac{1}{128} \, \left( (a_H / c) \, \tau Q_{HH} \right)^4 + \dots \right).$$
Together therefore:
$$\frac{\tau T_{PP}}{\tau P_{TT}} \approx \frac{\tau H_{QQ}}{\tau Q_{HH}} \approx 1 - \frac{1}{3} \, (a_H / c^2) \, QA + \mathcal O[ \, \left( (a_H / c^2) \, QA \right)^2 \, ],$$
where from $a_H = g_T \approx 9.81~{\rm m/s^2}$ and $QA \approx 1~{\rm km}$ obtains $(a_H / c^2) \, QA \approx 1.1 \times 10^{-13}$.
