In a more general setting than that of the original question: suppose we have a faithful normal state $h$ on a von Neumann algebra $M$. Suppose furthermore that $h$ is tracial, meaning that $h(xy)=h(yx)$ for all $x,y\in M$. (Warning! there are important examples of compact quantum groups where the Haar state is faithful but not tracial.)

In this setting we may define $\Vert x\Vert_{L^1(M,h)}$ to be $h(|x|)$. With this definition it is not clear that we have a norm; however, it is known that one has $$ h(|x|) = \sup\{ | h(xy) | \colon y \in M, \Vert y\Vert_M \leq 1 \} \tag{$*$}$$ and the proof has been given by Martin Argerami on MathStackExchange. The motivating example to keep in mind is the commutative case $M=L^\infty[0,1]$ with usual ess.sup norm and usual weak-star topology, with $h(x) = \int_0^1 x(t)\,dt$.

Historical note: if one looks at Definition 3.2 of I. Segal's paper

MR0054864 (14,991f) I. E. Segal, A non-commutative extension of abstract integration. Ann. of Math. (2) 57 (1953). 401–457

one sees that Segal took the RHS of $(*)$ as the definition of the $L^1$-norm on $M$ given by $h$. He then, in part (d) of Corollary 10.1, shows that the RHS of $(*)$ is equal to the LHS of $(*)$.

In a more general setting than that of the original question: suppose we have a faithful normal state $h$ on a von Neumann algebra $M$. Suppose furthermore that $h$ is tracial, meaning that $h(xy)=h(yx)$ for all $x,y\in M$. (Warning! there are important examples of compact quantum groups where the Haar state is faithful but not tracial.)

In this setting we may define $\Vert x\Vert_{L^1(M,h)}$ to be $h(|x|)$. With this definition it is not clear that we have a norm; however, it is known that one has $$ h(|x|) = \sup\{ | h(xy) | \colon y \in M, \Vert y\Vert_M \leq 1 \} \tag{$*$}$$ and the proof has been given by Martin Argerami on MathStackExchange. The motivating example to keep in mind is the commutative case $M=L^\infty[0,1]$ with usual ess.sup norm and usual weak-star topology, with $h(x) = \int_0^1 x(t)\,dt$.

Historical note: if one looks at Definition 3.2 of I. Segal's paper

MR0054864 (14,991f) I. E. Segal, A non-commutative extension of abstract integration. Ann. of Math. (2) 57 (1953). 401–457

one sees that Segal took the RHS of $(*)$ as the definition of the $L^1$-norm on $M$ given by $h$. He then, in part (d) of Corollary 10.1, shows that the RHS of $(*)$ is equal to the LHS of $(*)$.