The best constant is $C^{d-1}\sqrt{d}$.

Write $D(A)=\det A$. We can rephrase the inequality as the claim that $\|D'\|_F\le L$ for $c\le A\le C$. (It's perhaps best to think of the matrices as long column vectors and the Frobenius norm as the Euclidean norm, and then $D'$ can be viewed as the gradient of $D$.)

By [Jacobi's formula][1] for the derivative of a determinant, we have
$$
\frac{\partial D}{\partial a_{jk}}= D(A)\;\textrm{tr}\: (A^{-1}E_{jk}) = D(A) (A^{-1})_{kj} ;
$$
here $E_{jk}$ is the matrix with a $1$ in the $jk$ slot and zero entries otherwise.

So $\|D'(A)\|_F= D(A) \|A^{-1}\|_F$. (This immediately recovers the original inequality since $D\le C^d$, $\|A^{-1}\|^2_F\le d/c^2$.) To find the optimal constant, we write
$$
D^2\|A^{-1}\|^2_F= \prod \lambda_j^2\cdot \sum \lambda_j^{-2} = \lambda_2^2\cdots\lambda_d^2 +\lambda_1^2\lambda_3^2\cdots \lambda^2_d +\ldots +\lambda_1^2\cdots \lambda^2_{d-1}
$$
in terms of the eigenvalues of $A$. Clearly this is maximized at $A=C$, so the best constant is $C^{d-1}\sqrt{d}$.


  [1]: https://en.wikipedia.org/wiki/Jacobi%27s_formula