Analytic proof of Serre vanishing theorem Consider the following equivalent statement of Serre vanishing theorem (replacing ampleness condition on the line bundle with postivity condition).
Let $X$ be a compact complex  manifold. Let $L$ be a line bundle on $X$ admitting a hermitian metric with positive curvature form and let $F$ be a vector bundle on $X$. Then there exists $n_0$ such that for all $n \geq n_0$ and all $i > 0$
$$
H^i(X, F \otimes L^{\otimes n}) = 0
$$
(the fact that $F$ is a vector bundle allows us to think about $H^i$ as Dolbeault cohomology)
I wonder if it is possible to come up with an analytic proof of this statement, e.g. involving Hodge theory and Kähler identities, similarly to Kodaira vanishing theorem. Is such proof written down somewhere?
 A: You need $F$ to be a holomorphic vector bundle, but yes, an analytic proof exists. See Huybrechts' Complex Geometry: An Introduction, Proposition $5.2.7$. The proof is along the same lines as the analytic proof of the Kodaira vanishing theorem (which is Proposition $5.2.2$ in Huybrechts' book). 
Note, the statement of Proposition $5.2.7$ requires $X$ to be Kähler, which you do not, but the existence of a positive line bundle on $X$ implies it is Kähler, so it really is the same as the statement you provided.
I have written the proof below following the one given by Huybrechts. I have slightly changed the presentation and added some extra details. Note, below $m$ denotes the power of the line bundle and $n$ denotes the complex dimension of $X$ (Huybrechts uses the same notation), which differs from what you've written in your question.
Proof: Let $E$ be a holomorphic vector bundle. After choosing hermitian metrics on $E$ and $L$, we have the associated Chern connections $\nabla_E$ and $\nabla_L$. We then have the associated product connections $\nabla_{L^{\otimes m}}$ and $\nabla_{E\otimes L^{\otimes m}}$. By the hypothesis on $L$, we can assume the hermitian metric on $L$ is such that $\omega = i\nabla_L$ is a positive $(1, 1)$-form. By the first Bianchi identity, $\omega$ is parallel, so it is also closed. So we can equip $X$ with the Kähler metric whose associated $(1, 1)$-form is $\omega$.
A computation using the vector bundle analogues of the Kähler identities shows 
$$([\Lambda, iF_{\nabla_{E\otimes L^{\otimes m}}}]\alpha, \alpha) \geq 0$$
for all $\alpha \in \mathcal{H}^{p,q}(X, E\otimes L^{\otimes m})$ where the parentheses indicate the $L^2$ inner product on $(p, q)$-forms; see Lemma $5.2.4$ for details. Note that 
\begin{align*}
iF_{\nabla_{E\otimes L^{\otimes m}}} &= iF_{\nabla_E}\otimes 1 + 1\otimes iF_{\nabla_{L^{\otimes m}}}\\ 
&= iF_{\nabla_E}\otimes 1 + 1\otimes miF_{\nabla_L}\\ 
&= iF_{\nabla_E}\otimes 1 + m(1\otimes \omega).
\end{align*}
Therefore
\begin{align*}
([\Lambda, iF_{\nabla_{E\otimes L^{\otimes m}}}]\alpha, \alpha) &= ([\Lambda, iF_{\nabla_E}]\alpha, \alpha) + m([\Lambda, L]\alpha, \alpha)\\ 
&= ([\Lambda, iF_{\nabla_E}]\alpha, \alpha) + m(n-(p+q))\|\alpha\|^2
\end{align*}
where $L$ denotes the Lefschetz operator (not the line bundle).
Let a subscript $x$ denote the restriction to $\bigwedge^{p,q}T^*_xX$. Then $[\Lambda, iF_{\nabla_E}]_x$ is a linear operator acting on the vector space $\bigwedge^{p,q}T^*_xX$ equipped with the inner product $(\cdot, \cdot)_x$. By the Cauchy-Schwarz inequality, we have $|([\Lambda, iF_{\nabla_E}]_x\alpha_x, \alpha_x)_x| \leq \|[\Lambda, iF_{\nabla_E}]_x\alpha_x\|_x\|\alpha_x\|_x \leq \mathcal{C}_x\|\alpha_x\|_x^2$ where $\mathcal{C}_x$ denotes the operator norm of $[\Lambda, iF_{\nabla_E}]_x$. Let $\mathcal{C}$ denote the continuous function $X \to \mathbb{R}$, $x \mapsto \mathcal{C}_x$. As $\mathcal{C}$ is continuous and $X$ is compact, $\mathcal{C}$ attains a maximum, call it $C$. Then we have
\begin{align*}
|([\Lambda, iF_{\nabla_E}]\alpha, \alpha)| &= \left|\int_X([\Lambda, iF_{\nabla_E}]_x\alpha_x, \alpha_x)_x dV\right|\\ 
&\leq \int_X|([\Lambda, iF_{\nabla_E}]_x\alpha_x, \alpha_x)_x|dV\\ 
&\leq \int_X\mathcal{C}(x)\|\alpha_x\|_x^2dV\\
&\leq \int_X C\|\alpha\|_x^2dV\\
&= C\|\alpha\|^2.
\end{align*}
So we have 
\begin{align*}
0 \leq ([\Lambda, iF_{\nabla_{E\otimes L^{\otimes m}}}]\alpha, \alpha) &\leq ([\Lambda, iF_{\nabla_E}]\alpha, \alpha) + m(n-(p+q))\|\alpha\|^2\\
&\leq |([\Lambda, iF_{\nabla_E}]\alpha, \alpha)| + m(n-(p+q))\|\alpha\|^2\\
&= |([\Lambda, iF_{\nabla_E}]\alpha, \alpha)| + m(n-(p+q))\|\alpha\|^2\\
&\leq C\|\alpha\|^2 + m(n-(p+q))\|\alpha\|^2.
\end{align*}
Therefore, if $C + m(n - (p+q)) < 0$, $\alpha = 0$ so $\mathcal{H}^{p,q}(X, E\otimes L^{\otimes m}) = 0$. In particular, if $p = n$ and $q \geq 1$, then for any $m \geq m_0 := 1 + \lceil C\rceil$, we see that 
$$H^{n, q}_{\bar{\partial}}(X, E\otimes L^{\otimes m}) \cong \mathcal{H}^{n,q}(X, E\otimes L^{\otimes m}) = 0.$$ 
But by the Dolbeault Theorem, we have $H^{n,q}(X, E\otimes L^{\otimes m}) \cong H^q(X, K_X\otimes E\otimes L^{\otimes m})$ where $K_X = \bigwedge^nT^*X$ is the canonical line bundle. So we've shown that for any holomorphic vector bundle $E$, there is $m_0$ such that $H^q(X, K_X\otimes E\otimes L^{\otimes m}) = 0$ for all $q \geq 1$ and $m \geq m_0$. Now choose $E = K_X^*\otimes F$. Then there is an $m_0$ such that for all $q \geq 1$ and $m \geq m_0$, we have
$$0 = H^q(X, K_X\otimes E\otimes L^{\otimes m}) = H^q(X, K_X\otimes K_X^*\otimes F\otimes L^{\otimes m}) = H^q(X, F\otimes L^{\otimes m}).$$
