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The triangle circumcenter is conveniently expressed in trilinear coordinates as $$ \cos\alpha_1 : \cos\alpha_2 : \cos\alpha_3, $$ where the $\alpha_i$ are the angles opposite of the respective edge in the trilinear system.

Is there a similar expression for the circumcenter in quadriplanar coordinates, i.e., the same concept carried over to tetrahedra?

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    It is possible that this kind of question has an answer in some old documents, such as the "Mathematical Times" (https://archive.org/stream/mathematicalque03unkngoog#page/n12/mode/1up). I would advise you to look as well to barycentrical coordinates which are more common (because they have more properties). – 2017-02-14

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Let's consider the tetrahedron

enter image description here

and let's say we'd like to find out about the quadriplanar coordinate corresponding to point $A$. Let's further assume that the edges are directed like this: $$ a' = \overrightarrow{AD}, b = \overrightarrow{AC}, c = \overrightarrow{AB}\\ b' = \overrightarrow{BD}, c' = \overrightarrow{DC}, a = \overrightarrow{CB}. $$ Then the elevation of the circumcenter above the plane spanned by $C, B, D$ is given by $$ c_A = \\ (\langle a, b' \rangle \langle b', c' \rangle \langle c, a' \rangle +\\ \langle b', c' \rangle \langle c', a \rangle \langle a', b \rangle +\\ \langle c', a \rangle \langle a, b' \rangle \langle b, c \rangle +\\ \langle a, b' \rangle \langle b', c' \rangle \langle c', a \rangle )\\ / 24\\ /\alpha_{BDC}\\ / v, $$ where $v$ is the volume of the tetrahedron and $\alpha_{BDC}$ the area of the face $BDC$.

The same can be done for all other faces.

Since the quadriplanar coordinates are invariant under multiplication, the fixed factors $24$ and $v$ can be left aside.