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 $A=C$, with value $C^{d-1}\sqrt{d}$, so this constant works.

It is also optimal as we can confirm by simply taking $A=C$, $B=C-\epsilon$: then $\|A-B\|^2_F=d\epsilon^2$, $D(A)-D(B)=C^d-(C-\epsilon)^d=dC^{d-1}\epsilon + O(\epsilon^2)$.

(The fact that the derivative does have this value $C^{d-1}\sqrt{d}$ does not immediately rule out still smaller constants since the derivative controls movements in arbitrary directions while we are dealing with positive definite matrices only. However, we can also observe, in more abstract style, that $D'(A)$ is symmetric, so we do stay inside positive definite matrices when following the direction of the largest change of $D(A)$.)


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