It can be simplified as Tr$(\gamma^\nu \not{p_2}\gamma^\mu \gamma_\nu \not{p_1\gamma_{\mu}})$=Tr$(\gamma^\nu \gamma^{\rho}p_{2\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}p_{1\sigma}\gamma_{\mu})$.
Since $p$ doesn't carry spinor indices.
$p_{2\rho}p_{1\sigma}$Tr$(\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}\gamma_{\mu})$ using ${\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu}=4g^{\rho\mu}$
Tr$(\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}\gamma_{\mu})$= Tr$(4g^{\rho\mu} \gamma^{\sigma}\gamma_{\mu})$=$2$ Tr$( \{ \gamma^{\rho},\gamma^{\mu}\} \gamma^{\sigma}\gamma_{\mu})$
$\implies2$ Tr$( \gamma^{\rho}\gamma^{\mu}\gamma^{\sigma}\gamma_{\mu})$ $+$ $2$ Tr($\gamma^{\mu}\gamma^{\rho} \gamma^{\sigma}\gamma_{\mu})$=$2$Tr$(\gamma^{\rho}\times-2\gamma^{\sigma})$+$2$Tr$(4g^{\rho\sigma})$ where I have used $\gamma^{\mu}\gamma^{\sigma}\gamma_{\mu}=-2\gamma^{\sigma}$
Since $g^{\rho\sigma}$ doesn't have any spinor indicies we can take it out of second trace.
$\implies-4$Tr$(\gamma^\rho\gamma^\sigma)+8g^{\rho\sigma}$Tr$(\delta_{\alpha\beta})=-16g^{\rho\sigma}+64g^{\rho\sigma}$$(\delta_{\alpha\beta})=-16g^{\rho\sigma}+32g^{\rho\sigma}$
Orignal expression then becomes $p_{2\rho}p_{1\sigma}\times48g^{\rho\sigma}$$p_{2\rho}p_{1\sigma}\times16g^{\rho\sigma}$ =$48p_1\cdot p_2$$16p_1\cdot p_2$
