$$\require{cancel}$$
The kinetic energy is a scalar quantity and thus is the same in both stationary and rotating frames of reference. It is much easier to evaluate the kinetic energy in the rotating Principal-axis frame since the inertia tensor is diagonal in the Principal-axis frame as given in equation $$(13.12.14)$$
$T_{rot} = \frac{1}{2} \sum^3_i I_{ii} \omega^2_i$
Using equation $$(13.14.1-13.14.3)$$ for the body-fixed angular velocities gives the rotational kinetic energy in terms of the Euler angular velocities and principal-frame moments of inertia to be
$T_{rot}=\frac{1}{2}\left[I_{1}(\dot{\phi} \sin \theta \sin \psi+\dot{\theta} \cos \psi)^{2}+I_{2}(\dot{\phi} \sin \theta \cos \psi-\dot{\theta} \sin \psi)^{2}+I_{3}(\dot{\phi} \cos \theta+\dot{\psi})^{2}\right]$