Volume of a solid created by an extended tetrahedron 
Every edge of a tetrahedron with length $p$ is extended through the vertices by $p$.
Now all 12 points create a new solid $J$ of which I seek the volume dependent on the volume of the tetrahedron in the centre.


With some help the solution becomes clear:
The whole Volume of all pyramids is:
$$V_P=4\cdot\frac{1}{6}\cdot\left(\frac{a}{\sqrt{2}}\right)^3 +4\cdot\frac{1}{6}\cdot\left(\sqrt{2}a\right)^3 =\frac{3}{\sqrt{2}}a^3 \tag{1}$$
The side of the large cube is
$$s=\frac{3}{\sqrt{2}}a\tag{2}$$
and the Volume is respectively
$$V_C=\left(\frac{3}{\sqrt{2}}a\right)^3=\frac{27\sqrt{2}}{4}\cdot a^3 \tag{3}$$
The last step is subtracting the Volume of the pyramids.
$$V_J=V_C-V_P= \frac{27\sqrt{2}a^3}{4}-\frac{3}{\sqrt{2}}a^3=\frac{21\sqrt{2}a^3}{4} \tag{4}$$
Thx for the help.

Here is an answer - complementing your 3d object to a cube with side - $3p/2^{0.5}$

Subtract 4 right angle pyramids with diagonal $p$ and 4 right angle pyramids with diagonal $2p$. The pyramids sides are, $p/2^{0.5}$ and $2^{0.5}p$.
The volume of the cube is given by:$(3p/2^{0.5})^3$
The volume of the pyramids - $4X(1/6)X(p/2^{0.5})^3+4X(1/6)X(2p/2^{0.5})^3$
I will leave you to subtract the volumes.
