I belive that a approach by block matrix decompositon give a satisfactory answer. Seting
$ A=\left(\sum_{i=1}^{n}\sum_{i=1}^{n} c_{ij}\right)^2, \quad v^T= \left(\left(\sum_{j=1}^{n}c_{1j}\right)^2, \dots,\left(\sum_{j=1} c_{nj}\right)^2\right) $ $ v=\begin{pmatrix} \left(\sum_{j=1}^{n}c_{1j}\right)^2\\ \vdots \\ \left(\sum_{j=1}^{n}c_{nj}\right)^2 \end{pmatrix}, \quad B = \begin{pmatrix} c_{11}^2&\dots&c_{1j}^2&\dots&c_{1n}^2\\ \vdots& &\vdots& &\vdots \\ c_{i1}^2&\dots&c_{ij}^2&\dots&c_{in}^2\\ \vdots& &\vdots& &\vdots \\ c_{n1}^2 &\dots& c_{nj }^2&\dots&c_{nn}^2 \\ \end{pmatrix} $ We have $(v^TB)=(Bv)^T$ and $v^TBv\in\mathbb{R}$.
Suppose $A$, $D$, $C$, and $B$ are $n\times n$, $n\times m$, $m\times n$-, and $m\times m$ matrices, respectively. Then
$\det\begin{pmatrix}A& 0\\ C& B\end{pmatrix} = \det\begin{pmatrix}A& D\\ 0& B\end{pmatrix} = \det(A) \det(B) . $
This can be seen from the Leibniz formula for determinants or by induction on ''n''. When ''A'' is invertible matrix, employing the following identity
$ \begin{pmatrix}A& D\\ C& B\end{pmatrix} = \begin{pmatrix}A& 0\\ C& I\end{pmatrix} \begin{pmatrix}I& A^{-1} D\\ 0& B - C A^{-1} D\end{pmatrix}$
leads to
$\det\begin{pmatrix}A& D\\ C& B\end{pmatrix} = \det(A) \det(B - C A^{-1} D) .$
When ''B'' is invertible, a similar identity with $\det(B)$ factored out can be derived analogously,These identities were taken from http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/proof003.html that is,
$\det\begin{pmatrix}A& D\\ C& B\end{pmatrix} = \det(B) \det(A - D B^{-1} C) .$
When the blocks are square matrices of the same order further formulas hold. For example, if $C$ and $B$ commute (i.e., $CB = BC$), then the following formula comparable to the determinant of a 2-by-2 matrix holds:Proofs are given in J.R. Silvester, Math. Gazette, 10 (2000), pp. 460-467, available at http://www.mth.kcl.ac.uk/~jrs/gazette/blocks.pdf $ \det\begin{pmatrix}A& D\\ C& B\end{pmatrix} = \det(AB - DC).$
Them $\det(C)= \det(B)\cdot( A-v^TB^{-1}v)$, whit $v^TB^{-1}v, A\in\mathbb{R}$ or $ \det(C)= \det(B)\left[ \left(\sum_{i=1}^{n}\sum_{i=1}^{n} c_{ij}\right)^2 - \left(\left(\sum_{j=1}^{n}c_{1j}\right)^2, \dots,\left(\sum_{j=1} c_{nj}\right)^2\right) B^{-1} \begin{pmatrix} \left(\sum_{j=1}^{n}c_{1j}\right)^2\\ \vdots \\ \left(\sum_{j=1}^{n}c_{nj}\right)^2 \end{pmatrix} \right] $ whit $ \left(\left(\sum_{j=1}^{n}c_{1j}\right)^2, \dots,\left(\sum_{j=1} c_{nj}\right)^2\right) B^{-1} \begin{pmatrix} \left(\sum_{j=1}^{n}c_{1j}\right)^2\\ \vdots \\ \left(\sum_{j=1}^{n}c_{nj}\right)^2 \end{pmatrix} \in\mathbb{R}$.