I am finding some trouble in calculating the Linearized equation for even Laplacian. The $p$th Laplacian is defined as $-\Delta_pv= \text{div}(|Dv|^{p−2}Dv)$.
I know the linearized $p$ th Laplacian operator is $L_v(\phi)= -\text{div}(|Dv|^{p−2}|D\phi| + (p − 2)|Dv|^{p−4}|(Dv \cdot D\phi)Dv)$. I know the definition of it but I am unable to calculate $I'$ explicitly where $I$ is associated functional. It will be extremely helpful if one can give an idea how to calculate it explicitly or give some reference where such kind of calculations has been done.
First of all your definition of the $p$-Laplacian is not correct. The $p$-Laplacian is $$\tag{$\ast$}\Delta_pu(x) = \operatorname{div} (\vert D u \vert^{p-2} D u )$$ not $\operatorname{div} (\vert D u \vert^{p-2} \vert D u \vert )$ (the second doesn't make sense). You have a similar issue with the linearisation of the $p$-Laplacian. Also, you write $\Delta_p u=-\operatorname{div} (\vert D u \vert^{p-2} D u )$ instead of $(\ast)$ which is a non-standard sign convention---this is not a huge problem, but I will use the more standard definition $(\ast)$ in my answer.
To compute the linearisation of $\Delta_p$ around $v$ in the direction $\phi$ (denoted $L_v\phi$), you need to compute the power series expansion of $\Delta_p(v+\varepsilon \phi)$ as $\varepsilon\to0^+$. Then, by definition, $$\Delta_p(v+\varepsilon \phi) = \Delta_pv+ \varepsilon L_v\phi + O(\varepsilon^2) $$ using Big-O notation.
Now, we have that \begin{align*}\Delta_p (v + \varepsilon \phi) &= \operatorname{div} \big (\vert D v+\varepsilon D\phi \vert^{p-2} ( D v +\varepsilon D\phi )\big )\\ &= \operatorname{div} (\vert D v+\varepsilon D\phi \vert^{p-2} D v )+\varepsilon\operatorname{div} (\vert D v+\varepsilon D\phi \vert^{p-2} D\phi ) \end{align*} Next, \begin{align*} \vert D v+\varepsilon D\phi \vert^2&=\vert Dv\vert^2+2\varepsilon Dv\cdot D\phi + \varepsilon^2\vert D\phi\vert^2. \end{align*} Now, using that $(1+x)^\alpha=1+\alpha x +O(x^2)$ as $x\to 0$, we have that \begin{align*} \vert D v+\varepsilon D\phi \vert^{p-2} &= \big ( \vert Dv\vert^2+2\varepsilon Dv\cdot D\phi + \varepsilon^2\vert D\phi\vert^2\big )^{\frac{p-2}2}\\ &=\vert Dv\vert^{p-2} \bigg (1+2\varepsilon \frac{Dv\cdot D\phi}{\vert D v \vert^2} + O(\varepsilon^2) \bigg)^{\frac{p-2} 2 }\\ &=\vert Dv\vert^{p-2} \bigg (1+(p-2)\varepsilon \frac{Dv\cdot D\phi}{\vert D v \vert^2} + O(\varepsilon^2) \bigg)\\ &= \vert Dv\vert^{p-2} +(p-2)\varepsilon \vert Dv\vert^{p-4}Dv\cdot D\phi + O(\varepsilon^2). \end{align*} Hence, \begin{align*} &\hspace{-2em}\Delta_p (v + \varepsilon \phi) \\ &= \operatorname{div} (\vert D v\vert^{p-2} D v )+(p-2)\varepsilon\operatorname{div} (\vert Dv\vert^{p-4}(Dv\cdot D\phi )D v )+\varepsilon\operatorname{div} (\vert D v \vert^{p-2} D\phi ) +O(\varepsilon^2)\\ &=\Delta_pv + \varepsilon \operatorname{div}\big (\vert D v \vert^{p-2} D\phi +(p-2) \vert Dv\vert^{p-4}(Dv\cdot D\phi )D v \big ) +O(\varepsilon^2), \end{align*} which gives $$L_v\phi =\operatorname{div}\big (\vert D v \vert^{p-2} D\phi +(p-2) \vert Dv\vert^{p-4}(Dv\cdot D\phi )D v \big ). $$