Let us consider a standard discrete "spectrogram" or short-term Fourier transform with a window $w$ and frames of length $N$. Let $\|x\|_r = (\sum_{n=0}^{N-1}|x_n|^r)^{1/r}$ denote the $r$-norm of $x$ ($r\ge1)$. Then for each frame of $x$, the first windowed Fourier transform is:
$$F(\omega)=\sum_{n=0}^{N-1}x_n w_n e^{-j\omega n}\,.$$
Hence, by the Rogers-Hölder's inequalities, for all $\omega$, and any couple $(p,q)$ such that $\frac{1}{p}+\frac{1}{q}=1$:
$$|F(\omega) |\le \|w\|_p\|x\|_q\,.$$
In particular, for classical positive unit windows, $\sum_{n=0}^{N-1} w_n =1 = \|w\|_1$. Choose $p=1$ and $q=\infty$, and you get (over all frames):
$$|F(\omega) | \le \max_n | x_n|\,.$$
Not the tighest upper bound, but simple enough. And honestly, using Rogers-Hölder was seen a bit of an overkill here, since simply:
$$\left|\sum_{n=0}^{N-1}x_n w_n e^{-j\omega n} \right| \le \sum_{n=0}^{N-1}\left |x_n w_n e^{-j\omega n} \right| \le \max_n | x_n|\sum_{n=0}^{N-1}\left | w_n \right|\,, $$
but for "unit-energy" windows, you will get other estimates with $p=q=2$.
You can now plug it inside your "log of one plus" formula.
Similarly, you could find a tighter lower bound.