Let $A$ be a real square matrix with non-zero determinant. Define the quantity $C(A)$ by $$C(A) = d! \sum_{i=0}^{d-1} {d\choose i} \|A\|_F^i.$$ Consider now the ratio $B(A) = C(A)^2/|\det A|$.
General question Is $B(A)$ bounded from above? This is perhaps too general, so let's think about a more specific case.
Specific case Suppose $A=D\phi^T D\phi$ for some smooth function $\phi$ on an openset $U\subset \mathbb{R}^d$, where $D\phi$ is the Jacobian matrix. Further suppose that $\|D\phi(u)\|_F\leq 1$ for all $u$ and that $\det A(u)\neq 0$. Can we bound $B(A(u))$ in this case independent of $u$?
So far, we can note that $\|A(u)\|_F\leq \|D\phi(u)\|_F^2$ by sub-multiplicativity of the Frobenius norm. Therefore by the hypothesis that the Jacobian is bounded by $1$ in the Frobenius norm, we have that $$C(A) \leq d! \sum_{i=0}^{d-1} {d\choose i} \leq d! \left(\frac{ed}{d-1}\right)^{d-1}$$ $$ \leq d^d e^{d-1}.$$ Okay, that takes care of the numerator. Now we need a lower bound for $ |\det A(u)|$. We could get one in terms of eigenvalues but these will depend on $u$. If these eigenvalues were continuous over $u$, we could work on a compact subset $K$ of $U$ and take the minimum of such over $K$. Since eigenvalues are roots of a polynomial, and such roots are continuous with respect to the coefficients, this seems fine. But is there anything else we could do that I am not seeing?
No, there is no such bound. The precise functional form of $C(A)$ is not really relevant; the point is that we can make the Frobenius norm arbitrarily large compared to the determinant.
Actually we can already do this with a diagonal matrix; take $A = \text{diag}(1, 1, \dots 1, \varepsilon)$, which has Frobenius norm $\sqrt{(d-1) + \varepsilon^2}$ but determinant $\varepsilon$, and then take $\varepsilon \to 0$. This matrix is even symmetric and positive-definite if $\varepsilon > 0$ so it works for your more specific case too. To make the Frobenius norm $\le 1$ we can just divide by $\sqrt{d}$ which actually makes the situation significantly worse since the determinant is now $\varepsilon \sqrt{d}^{-d}$.
Even without this observation we can notice what happens when we scale $A$ by a scale factor $\lambda > 0$. If these are $d \times d$ matrices (which you haven't specified but which I'm assuming based on your example) then $|\det(A)|$ is scaled by $\lambda^d$ but $\| A \|_F^i$ is scaled by $\lambda^i$. The worst term in the numerator is actually the constant term $d!$ which doesn't scale at all. This means that $C(A) \ge d!$ regardless of how the other terms behave, and meanwhile by scaling $A$ by a very small $\lambda > 0$ both the denominator and the other terms in the numerator become arbitrarily small.