The operator norm of $T$ is the supremum over all unit vectors $v$ of the quantity $\langle Tv, Tv \rangle^{1/2}$. However, for positive self-adjoint operators we may replace the quantity by $\langle Tv, v \rangle$.
Does this work for any other operators besides positive self-adjoint ones?
For any bounded operator $T$ on a Hilbert space $\mathcal H$, one has $$ \lVert T \rVert=\sup \lVert Tv\rVert = \sqrt{\sup \langle Tv, Tv\rangle} = \sqrt{\sup \langle v, T^* Tv\rangle}, $$ where the suprema are over $v\in\mathcal{H}$ with $\lVert v \rVert = 1$.
If $T$ is self-adjoint, i.e. if $T^*=T$, then one has $$ \lVert T \rVert= \sup \lvert \langle v, Tv\rangle \rvert,\tag{1} $$ as shown in this answer. $T$ is called anti self-adjoint if $T^*=-T$, which is equivalent to $iT$ being self-adjoint. In this case we conclude that (1) holds true as well. Of course, if $T\geqslant 0$, formula (1) simplifies to $$ \lVert T \rVert= \sup \langle v, Tv\rangle. $$ In order for $\sup \langle v, Tv\rangle$ to make sense, we must have $\langle v, Tv\rangle \in \mathbb{R}$, which implies that $T$ is self-adjoint. In this case, we have $$ \sup \langle v, Tv\rangle = \max_{\lambda\in\sigma(T)} \lambda, $$ which may or may not coincide with $\lVert T \rVert$.
If one only assumes that $T$ is normal, i.e. that $T^*T=TT^*$, then formula (1) still holds true. This can be proven by means of the spectral theorem, see Theorem 12.25 in the book 'Functional Analysis' by Rudin.
In general, one has $$ \lVert T \rVert \leqslant 2 \sup \lvert \langle v, Tv\rangle \rvert.\tag{2} $$ In order to obtain this bound, note that one can decompose $T$ into a sum of a self-adjoint operator and an anti self-adjoint operator $$ T=(T+T^*)/2 + (T-T^*)/2. $$ But this implies \begin{align*} \lVert T \rVert &\leqslant \lVert T+T^* \rVert /2 + \lVert T-T^* \rVert /2 \\ &= \sup\lvert \langle v,Tv\rangle + \langle Tv,v\rangle \rvert/2 + \sup\lvert \langle v,Tv\rangle - \langle Tv,v\rangle \rvert/2 \\ &\leqslant 2 \sup\lvert \langle v,Tv\rangle \rvert. \end{align*}
Finally, inequality (2) is optimal, as one may readily verify by considering the $2\times 2$ matrix $$ T=\left[\begin{matrix} 0 & 1 \\ 0 & 0\end{matrix}\right]. $$