In the middle of Page 42 of Ballmann's book, the author defines the Hermitian metric by $h=g+i\omega$, where $g$ is the compatible Riemannian metric which is a Riemannian metric $g$ satisfying $g(X,Y)=g(JX,JY),\forall X,Y\in T_{\mathbb C}M$ and $\omega:=g(JX,Y)$ is the associated fundamental form of $g$.
Then in the next Page (p.43), 4.10 Examples. 1), the author says for the complex manifold $\mathbb C$, the Euclidean metric is $g=\frac{1}{2}dz\odot d\bar z=\frac{1}{2}(dz\otimes d\bar z+d\bar z\otimes dz)$ and $\omega=\frac{i}{2}dz\wedge d\bar z=\frac{i}{2}(dz\otimes d\bar z-d\bar z\otimes dz)$.
In this case, the Hermitian metric $h=g+i\omega=\frac{1}{2}(dz\otimes d\bar z+d\bar z\otimes dz)-\frac{1}{2}(dz\otimes d\bar z-d\bar z\otimes dz)=d\bar z\otimes dz$, does it mean the standard Hermitian metric of $\mathbb C$ is $d\bar z\otimes dz$?
Then how to evaluate the length of $\frac{\partial }{\partial z}$, or how to compute $|\frac{\partial }{\partial z}|^2$?
If $|\frac{\partial }{\partial z}|^2$ is defined by $h(\frac{\partial }{\partial z},\frac{\partial }{\partial \bar z})$, does it mean the length is zero?
By the way, how to compute $|\frac{\partial }{\partial \bar z}|^2$?
Note Ballmann's convention for Hermitian metrics (quoting from page $1$):
With this convention, your calculation is accurate. If one wishes to use the convention that Hermitian metrics are complex-linear in the first variable instead, then the standard Hermitian product on $\mathbb{C}^n$ is $\mathrm{d}z\otimes\mathrm{d}\bar{z}=g-\mathrm{i}\,\omega$.
So for now we will use this definition: a Hermitian metric on a complex vector space $V$ satisfies $h(v_1,v_2)=\overline{h(v_2,v_1)}$ and $h(\lambda\,v_1,\mu\,v_2)=\bar{\lambda}\mu\,h(v_1,v_2)$. Now, from any complex vector space $V$ you can get a real vector space $V_{\mathbb{R}}$ which as a set is still $V$, but with scalar multiplication restricted to the real numbers. The complex multiplication on $V$ induces an endomorphism $J$ of $V_{\mathbb{R}}$ satisfying $J^2=-\mathrm{Id}$. From $V_{\mathbb{R}}$ we can define a new complex vector space, $V_{\mathbb{C}}:=V_{\mathbb{R}}\otimes\mathbb{C}$, and we extend $J$ complex-linearly to obtain an endomorphism of $V_{\mathbb{C}}$.
We can split $V_{\mathbb{C}}$ as a direct sum of the eigenspaces for $J$ corresponding to the eigenvalues $\pm\,\mathrm{i}$, obtaining $V_{\mathbb{C}}=V^{1,0}\oplus V^{0,1}$. Now, what is the relation between the original $V$ and these new spaces? You can check that, as complex vector spaces, $V\cong V^{1,0}$. The isomorphism is given by the map $$v\mapsto \frac{1}{2}\left(v-\mathrm{i}\,J\,v\right),$$ with inverse $$v\mapsto\mathrm{Re}(v)=\frac{1}{2}(v+\bar{v})\in V_{\mathbb{R}}.$$ Let's get to the case $V=\mathbb{C}^n$. With the usual notation, $\partial_{z}$ is a section of $V^{1,0}$, while $\partial_{\bar{z}}$ is in $V^{0,1}$. On the face of it, it does not make much sense to plug $\partial_{z}$ in the Hermitian product $h$, which is defined on $V$. The notation $\lvert\partial_{z}\rvert^2$ usually stands for $$\lvert\partial_{z}\rvert^2=g(\partial_{z},\partial_{\bar{z}})=\lvert\partial_{\bar{z}}\rvert^2.$$ It is easy to check that $g(\partial_{z},\partial_{z})=g(\partial_{\bar{z}},\partial_{\bar{z}})=0$.
So, in which sense $h=\mathrm{d}\bar{z}\otimes\mathrm{d}z$? This identity holds on the extended vector space $V_{\mathbb{C}}$, on which however $h$ is not a Hermitian product. Anyway, we can relate the Hermitian product on $V$ with the pairing on $V_{\mathbb{C}}$: $g(\partial_z,\partial_{\bar{z}})=\frac{1}{2}\overline{h(\partial_x,\partial_x)}=\frac{1}{2}h(\partial_x,\partial_x)$ (see Remark $4.9$ in Ballmann's book), so the "length" is positive as you'd expect.
Now, let's look at some computations. The expression for $h$ on $V_{\mathbb{C}}$ in coordinates shows that $h(\partial_z,\partial_{\bar{z}})=0$ and $h(\partial_{\bar{z}},\partial_z)=1$, but we can also check this without using the differentials $\mathrm{d}z$ and $\mathrm{d}\bar{z}$. Take as $\mathbb{C}$-basis for $\mathbb{C}$ the vector $\partial_x$ (where $x$ is the usual real coordinate), and let $\partial_y=\mathrm{i}\,\partial_x=:J\partial_x$. Then $\partial_x,\partial_y$ is an $\mathbb{R}$-basis for $V_{\mathbb{R}}$, and $\partial_z=\frac{1}{2}(\partial_x-\mathrm{i}\,\partial_y)$ is a $\mathbb{C}$-basis for $V^{1,0}$. Together with $\partial_{\bar{z}}$, it forms a $\mathbb{C}$-basis for $V_{\mathbb{C}}$. If we extend $g$ and $\omega$ to $V_{\mathbb{C}}$ complex-linearly, also $h$ is extended, so it makes sense to compute $h(\partial_z,\partial_{\bar{z}})$. As $h$ is now complex-linear in $V_{\mathbb{C}}$, we get $$h(\partial_z,\partial_{\bar{z}})=\frac{1}{4}h\left(\partial_x-\mathrm{i}\,\partial_y,\partial_x+\mathrm{i}\,\partial_y\right)=\frac{1}{4}\left(h(\partial_x,\partial_x)-\mathrm{i}\,h(\partial_y,\partial_x)+\mathrm{i}\,h(\partial_x,\partial_y)+h(\partial_y,\partial_y)\right).$$ Now, $\partial_x$ and $\partial_y=J\partial_x$ are elements of $V_{\mathbb{R}}=V$ (as sets). Since $J$ on $V_{\mathbb{R}}$ is the multiplication by $\mathrm{i}$ coming from $V$, using that $h$ is Hermitian in $V$ we get $$h(\partial_z,\partial_{\bar{z}})=\frac{1}{4}\left(h(\partial_x,\partial_x)-\mathrm{i}\,h(\mathrm{i}\partial_x,\partial_x)+\mathrm{i}\,h(\partial_x,\mathrm{i}\partial_x)+h(\mathrm{i}\partial_x,\mathrm{i}\partial_x)\right)=\frac{1}{4}\left(h(\partial_x,\partial_x)+(-\mathrm{i})^2\,h(\partial_x,\partial_x)+\mathrm{i}^2\,h(\partial_x,\partial_x)+(-\mathrm{i})\mathrm{i}\,h(\partial_x,\partial_x)\right)=0.$$ The analogous computation with $\partial_z$ and $\partial_{\bar{z}}$ switched will give you $1$.