Norm of tensors on a Hermitian manifold On a Riemannian manifold $(M, g)$, given a $(l, k)$ tensor $T  = T_{m_1 \ldots m_{l}} dx^{1} \otimes \ldots \otimes dx^{l}$, we know its norm is given by $g^{m_1 n_1} \ldots g^{m_l n_l} T_{m_1 \ldots m_{l}} T_{n_1 \ldots n_{l}}$. If instead $(M, g)$ is a Hermitian manifold with an $(l, l)$ tensor $T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} dz^{m_1} \otimes d \bar z^{p_1} \otimes \ldots \otimes dz^{m_l} \otimes d \bar z^{p_l}$, the norm then will be $g^{m_1 \bar{q_1}} \ldots g^{m_n \bar{q_n}} T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} \overline{T_{n_1 \bar {q_1} \ldots n_{l} \bar q_{l}}}$.
This confuses me because if we are to imitate the definition in the real case we would want $g^{m_{1} n_{1}}g^{\bar p \bar q_1} \ldots T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} T_{n_1 \bar {q_1} \ldots n_{l} \bar q_{l}}$. However apparently this does not make sense beacuse $g^{m_1 n_1} = 0$.
Maybe in order to compare it with the real case one needs to transform the coordinates into real coordinates?
 A: Part of your confusion may be down to some of the definitions involved, so let me clear up a few things.

"On a Riemannian manifold $(M,g)$, given a $(l,k)$ tensor $T = T_{m_1\dots m_l} dx^1\otimes\cdots\otimes dx^l$"

That is not a $(l, k)$-tensor, that is a $(0, l)$-tensor. Likewise, on a complex manifold, $T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} dz^{m_1} \otimes d \bar z^{p_1} \otimes \cdots \otimes dz^{m_l} \otimes d \bar z^{p_l}$ is not an $(l, l)$-tensor, but rather a $(0, 2l)$-tensor.
If $T = T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\cdots\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\cdots\otimes d\bar{z}^{p_l}$ is a $(0, k + l)$ tensor on a hermitian manifold, then the square of the norm of $T$ is given by
$$g^{m_1\bar{n_1}}\cdots g^{m_k\bar{n_k}}g^{q_1\bar{p_1}}\cdots g^{q_l\bar{p_l}}T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{T_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}}.$$
This follows from the definition of the inner product $\langle\cdot, \cdot\rangle$ on $(0, k + l)$-tensors induced by the hermitian metric. You're viewing a hermitian metric as a complex bilinear pairing $g$ on $TM\otimes_{\mathbb{R}}\mathbb{C}$, so one obtains a metric on $T^{1,0}M$ by considering the pairing $(v, w) \mapsto g(v, \overline{w})$; note, that $g(\overline{w}, v) = g(v, \overline{w})$ as $g$ is symmetric. The pairing $g$ induces pairings on $T^*M\otimes_{\mathbb{R}}\mathbb{C}$ and $\bigotimes^{k+l}_{\mathbb{C}}T^*M\otimes_{\mathbb{R}}\mathbb{C}$ which I will also denote by $g$. With this in mind, we have
\begin{align*}
& \small{\langle T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{p_l}, S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}dz^{n_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{n_k}\otimes d\bar{z}^{q_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{q_l}\rangle}\\
=&\ \small{g(T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{p_l}, \overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}dz^{n_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{n_k}\otimes d\bar{z}^{q_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{q_l}})}\\
=&\ \small{g(T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{p_l}, \overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}}d\bar{z}^{n_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{n_k}\otimes dz^{q_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{q_l})}\\
=&\ \small{T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}} g(dz^{m_1}, d\bar{z}^{n_1})\cdots g(dz^{m_k}, d\bar{z}^{n_k})g(d\bar{z}^{p_1}, dz^{q_1})\cdots g(d\bar{z}^{p_l}, dz^{q_l})}\\
=&\ \small{T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}} g^{m_1\bar{n_1}}\cdots g^{m_k\bar{n_k}}g(dz^{q_1}, d\bar{z}^{p_1})\cdots g(dz^{q_l}, d\bar{z}^{p_l})}\\
=&\ \small{T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}} g^{m_1\bar{n_1}}\cdots g^{m_k\bar{n_k}}g^{q_1\bar{p_1}}\cdots g^{q_l\bar{p_l}}.}
\end{align*}
