As an answer by ja72, whom I heartily thank, confirms, my following calculations prove that $\frac{d\boldsymbol{\omega}}{dt}=I^{-1}\big(\frac{d\mathbf{L}_{cm}}{dt}-\boldsymbol{\omega}\times(I\boldsymbol{\omega})\big)$.
By using a moving Cartesian coordinate system, dextrogyre as usual when calculating angular quantities, with basis $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ solidally moving with the rigid body, whose vectors I use as the columns of a matrix $E(t)=\left( \begin{array}{ccc}\mathbf{i}(t)&\mathbf{j}(t)&\mathbf{k}(t) \end{array} \right)$ for the sake of brevity, we have that$$I=EI_mE^{-1},\quad\boldsymbol{\omega}=E\boldsymbol{\omega}_{m},\quad I\boldsymbol{\omega}=EI_m\boldsymbol{\omega}_{m}$$where I have used the index $m$ to mark the inertia matrix and, respectively, the triplet of the coordinates of angular velocity with respect to basis $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$. By differentiating I get$$\frac{d\mathbf{L}_{cm}}{dt}=\frac{dE}{dt}I_m\boldsymbol{\omega}_{m}+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}$$and, by taking Poisson formula, which give us $\frac{d}{dt}\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)=\left( \begin{array}{ccc}\boldsymbol{\omega}\times\mathbf{i}&\boldsymbol{\omega}\times\mathbf{j}&\boldsymbol{\omega}\times\mathbf{k} \end{array} \right)$, into account and calling $I_m^{(i)}$ the $i$-th row of $I_m$,$$\frac{d\mathbf{L}_{cm}}{dt}=\boldsymbol{\omega}\times( I_m^{(1)}\cdot\boldsymbol{\omega}_m\mathbf{i})+\boldsymbol{\omega}\times( I_m^{(2)}\cdot\boldsymbol{\omega}_m\mathbf{j})+\boldsymbol{\omega}\times( I_m^{(3)}\cdot\boldsymbol{\omega}_m\mathbf{k})+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}$$$$=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}$$and since $I_m$ does not depend upon time$$\frac{d\mathbf{L}_{cm}}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+EI_m\frac{d\boldsymbol{\omega}_{m}}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+I E\frac{d(E^{-1}\boldsymbol{\omega})}{dt}$$$$=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+I\bigg(\frac{d(\boldsymbol{\omega}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{k})}{dt}\mathbf{k}\bigg)$$but Poisson formulae give $\frac{d\mathbf{r}}{dt}=\frac{d(\mathbf{r}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\mathbf{r}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\mathbf{r}\cdot\mathbf{k})}{dt}\mathbf{k}+\boldsymbol{\omega}\times\mathbf{r}$ for any differentiable vector function and therefore $\frac{d\boldsymbol{\omega}}{dt}=\frac{d(\boldsymbol{\omega}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{k})}{dt}\mathbf{k}+\boldsymbol{\omega}\times\boldsymbol{\omega}$ where $\boldsymbol{\omega}\times\boldsymbol{\omega}=\mathbf{0}$, which give$$\frac{d\boldsymbol{\omega}}{dt}=I^{-1}\bigg(\frac{d\mathbf{L}_{cm}}{dt}-\boldsymbol{\omega}\times(I\boldsymbol{\omega})\bigg)$$where all the quantities are calculated with respect to the external inertial frame, and where $I=EI_mE^\text{T}$ with $E^{\text{T}}$ as the transpose matrix of $E$.
