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A portion of a wooden cube is sawed off at each vertex so that a small equilateral triangle is formed at each corner with vertices on the edges of the cube. The $24$ vertices of the new object are all connected to each other by straight lines. How many of these lines (with the exception, of course, of their end-points) lie entirely in the interior of the original cube?

Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:

$hspace{3cm}$

First floor: Number $1$ can connect with none on the first floor. $1$ can connect with ${11,12}$ on the second floor, with ${15,16}$ on the third floor and with ${19,20,21,22,23,24}$ on the fourth floor. So, $1$ can connect with $2+2+6=10$ points, so does each of $2-8$, hence: $8cdot 10=color{red}{80}$.

Second floor: Number $9$ can connect with none on the first floor. $9$ can connect with ${11}$ on the second floor, so does $10$ with $12$. Hence: $color{red}{2}$ connections on the second floor only. $9$ can connect with ${15}$ on the third floor and with ${19,20,21,22}$ on the fourth floor. So, $9$ can connect with $1+4=5$ points on the third and fourth floors, so does each each of ${10,11,12}$, hence: $4cdot 5=color{red}{20}$. Overall, $color{red}{22}$ connections.

Third floor: $13$ can connect with none on the first or second floor. $13$ can connect with ${15}$ on the third floor, so does $14$ with $16$, hence $color{red}{2}$ connections on the third floor only. $13$ can connect with ${19,20,21,22}$ on the fourth floor. So, $13$ can connect with $4$ points on the fourth floor, so does each of ${14,15,16}$, hence: $4cdot 4=color{red}{16}$. Overall, $color{red}{18}$ connections.

Fourth floor: $17$ can connect with none on the first, second, third or fourth floor. So does each other on the fourth floor.

Hence, the number of inner connections is: $color{red}{80}+color{red}{22}+color{red}{18}=120$.

In the resulting truncated cube, each vertex $v$ is incident to one triangle and two octagons, which together have $3+8+8-3-3=13$ vertices between them excluding $v$. Lines from $v$ to these $13$ vertices will not be completely in the cube; lines to the other $24-13-1=10$ vertices will. Since each connecting line is incident to two vertices, there are $frac{24cdot10}2=120$ interior lines.

Every one of them that does not run along a face.

Each octahedral face of the object, of which there are $6,$ contains $8$ vertices, so there are $6{8 choose 2}=168$ segments along those faces. The segments that are part of the $12$ original edges of the cube are counted twice, so we subtract $12$, leaving $156$ segments on the octagonal faces of the object.

The triangular faces have segments running along them, but we have already counted those because they also run along an octahedral face. There are ${24 choose 2}=276$ line segments, so there are $276-156=120$ running through the body of the object.

The object you describe is called the truncated cube, it's an Archimedian solid.
See for example:
https://en.wikipedia.org/wiki/Truncated_cube

All of its edges lie on the surface of the orginal cube.