Euler’s Equations of Motion for a Rigid Body and Inertial Forces

Do we then not need to take into account inertial forces when computing the torques in rotating coordinates?

No, but there is an inertial torque you have to worry about.

From the perspective of an inertial frame, the rotational analog of Newton's second law for rotation about the center of mass is $$frac{dboldsymbol L}{dt} = sum_i boldsymbol tau_{text{ext},i}tag{1}$$ where $boldsymbol L$ is the object's angular momentum with respect to inertial, $boldsymbol tau_{text{ext},i}$ is the $i^text{th}$ external torque, and the differentiation is from the perspective of the inertial frame. Note that this pertains to non-rigid objects as well as rigid bodies.

The relationship between the time derivatives of any vector quantity $boldsymbol q$ from the perspectives of co-located inertial and rotating frames is $$left(frac{dboldsymbol q}{dt}right)_text{inertial} = left(frac{dboldsymbol q}{dt}right)_text{rotating} + boldsymbol Omega times boldsymbol qtag{2}$$ where $boldsymbolOmega$ is the frame rotation rate with respect to inertial.

For a rigid body, the body's angular momentum with respect to inertial but expressed in body-fixed coordinates is $boldsymbol L = mathbf I,boldsymbol omega$ where $mathbf I$ is the body's moment of inertia tensor and $boldsymbol omega$ is the body's rotation rate with respect to inertial but expressed in body-fixed coordinates. Since a rigid body's inertia tensor is constant in the body-fixed frame, we have $$left(frac{dboldsymbol L}{dt}right)_text{body-fixed} = frac{d(mathbf I boldsymbol omega)}{dt} = mathbf I frac{dboldsymbolomega}{dt}tag{3}$$

Combining equations (1), (2), and (3) yields $$mathbf I frac{dboldsymbolomega}{dt} = sum_i boldsymbol tau_{text{ext},i} - boldsymbol omegatimes(mathbf I times boldsymbol omega)tag{4}$$

This is Euler's equations of motion for a rigid body. No inertial forces come into play. However, the term $-boldsymbol omegatimes(mathbf I times boldsymbol omega)$ is essentially an inertial torque. Just as inertial forces vanish in inertial frames, so does this inertial torque.

Euler's equations for a rigid body express the change in angular momentum with respect to the rotating system, but the torques are expressed for the inertial system. For example, see the explanations in Goldstein, Classical Mechanics or in Symon, Mechanics. This was confusing to me in my first Mechanics physics course.