So you have an object whose principal mass moment of inertia values $I_1$, $I_2$ and $I_3$ you know for the body-centered coordinates
Now you place this object on a tilted axis and rotate it about z
and you want to find out the mass moment of inertia $I_{zz}$ about z.
You correctly identified the rotational speed on the body coordinates as
$$ \begin{aligned}
\omega_2 &= \omega \sin \theta \\
\omega_3 &= \omega \cos \theta
\end{aligned}$$
and that these cause the following components of angular momentum
$$ \begin{aligned}
L_2 &= ( I_2 \sin \theta) \omega \\
L_3 &= (I_3 \cos \theta ) \omega
\end{aligned}$$
except you went with energy instead of momentum, which might have obscured things because energy is a single scalar, and momentum is a vector and thus easier to understand what the terms represent (what orientation things are defined in).
The last step is to rotate the momentum components back into the world coordinate system
$$ \begin{aligned}
L_{yz} &= L_2 \cos \theta - L_3 \sin \theta \\
L_{zz} &= L_2 \sin \theta + L_3 \cos \theta
\end{aligned}$$
which leads to the value of MMOI once the rotational speed is factored out
$$ \begin{aligned}
L_{y} &= \underbrace{\left( (I_2-I_3)\sin \theta \cos\theta \right)}_{I_{yz} } \omega \\
L_{z} &= \underbrace{\left( I_2 - (I_2-I_3)\cos^2\theta \right)}_{I_{zz} } \omega
\end{aligned}$$
So your answer is
$$ \boxed{ I_{zz} = I_2 - (I_2-I_3)\cos^2\theta } $$
But it is a lot easier for me to do the transformation using the following matrix product, because it will produce all the components of the mass moment of inertia tensor.
Given the body inertia values as the following diagonal 3×3 matrix $$I_{\rm body} = \pmatrix{I_1 & & \\ &I_2 & \\ & & I_3} $$
and the rotation about the x-axis by the angle $\theta$ as a 3×3 tranformation matrix $$ R = \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta} $$
then the mass moment of inertia tensor in the world coordinates is given by the congruent transformation $I_{\rm world} = R I_{\rm body} R^\top$
$$ I_{\rm world} = \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta}\pmatrix{I_1 & & \\ &I_2 & \\ & & I_3} \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta}^\top $$
$$ \boxed{ I_{\rm world} = \pmatrix{ I_1 & 0 & 0 \\
0 & I_3 + (I_2-I_3) \cos^2 \theta & (I_2-I_3)\sin \theta \cos \theta \\
0 & (I_2-I_3) \sin \theta \cos \theta & I_2+(I_3-I_2)\cos^2 \theta } }$$
In general, the rotational inertia of a rigid body is fully characterized by an inertia tensor, which has 6 independent variables: $I_{xx}$, $I_{yy}$, $I_{zz}$, $I_{xy}$, $I_{yz}$, $I_{zx}$. The three diagonal terms can be calculated from moments of inertia about the x,y,z axes (which are $I_x$, $I_y$, $I_z$ in your notation):
$I_{xx} = \frac{I_y + I_z - I_x}{2}$; $I_{yy} = \frac{I_z + I_x - I_y}{2}$; $I_{zz} = \frac{I_x + I_y - I_z}{2}$
Unfortunately, the three non-diagonal terms $I_{xy}$, $I_{yz}$, $I_{zx}$ cannot be calculated from $I_x$, $I_y$, $I_z$; they are independent quantities that must be measured separately. However, these terms are often zero when the object is symmetric: if there is reflective symmetry in the yz-plane, then $I_{xy} = I_{zx} = 0$; if there is reflective symmetry in the zx-plane, then $I_{yz} = I_{xy} = 0$. I suspect both of these are true for a human arm, to a good enough approximation for your purposes.
Once you know all the components of the inertia tensor, you can calculate the moment $I_n$ about an arbitrary unit vector $ \begin{pmatrix}
n_x\\
n_y\\
n_z
\end{pmatrix}$ using
$I_n = (I_{xx} + I_{yy} + I_{zz}) - I_{nn}$, where:
\begin{equation*}
I_{nn} =
\begin{pmatrix}
n_x\\
n_y\\
n_z
\end{pmatrix}^T
\begin{pmatrix}
I_{xx}&I_{xy}&I_{xz}\\
I_{yx}&I_{yy}&I_{yz}\\
I_{zx}&I_{zy}&I_{zz}
\end{pmatrix}
\begin{pmatrix}
n_x\\
n_y\\
n_z
\end{pmatrix}
\end{equation*}
Note: the tensor is always symmetric in the mathematical sense (even if the object has no geometric symmetry). This means the written order of the indices doesn't matter: $I_{xy} = I_{yx}$, etc. That is why the inertia tensor has 6 degrees of freedom, rather than 9.
Best Answer
The inertia $I$ is actually a tensor whose components are
$$ I_{ij} = \int{\rm d}^3{\bf x}~\rho({\bf x}) [{\bf x}\cdot{\bf x}\delta_{ij} - x_ix_j] \tag{1} $$
So, for example the component $I_{11}$ can be calculated as
$$ I_{11} = \int{\rm d}^3{\bf x}~\rho({\bf x}) [x^2 + y^2 + z^2 -x^2 ] = \int{\rm d}^3{\bf x}~\rho({\bf x}) [y^2 + z^2] \tag{2} $$
To calculate this we need the density, which for this problem is just
$$ \rho({\bf x}) = \rho(r,\phi,z) = \frac{M}{2\pi R h}\delta(r-R) \tag{3} $$
Replacing (3) in (2) you get
\begin{eqnarray} I_x &\stackrel{\rm def.}{=}& I_{11} = \int {\rm d}r {\rm d}\phi {\rm d}z ~r \left[ \frac{M}{2\pi R h}\delta(r-R) \right] (y^2 + z^2), ~~y=r\sin\phi \\ &=& \frac{M}{2\pi R h}\left\{ \int {\rm d}r {\rm d}\phi {\rm d}z ~r \delta(r-R)(r^2\sin^2\phi) + \int {\rm d}r {\rm d}\phi {\rm d}z ~r \delta(r-R)(z^2) \right\} \\ &=& \frac{M}{2\pi R h}\left\{ R^3h \int_0^{2\pi}{\rm d}\phi~\sin^2\phi + 2\pi R \int_{-h/2}^{h/2}{\rm d}z~z^2 \right\} \\ &=&\frac{M}{2\pi R h}\left\{ \pi R^3h + 2\pi R \frac{h^3}{12}\right\} \\ &=& \frac{M}{2}\left(R^2 + \frac{h^2}{6}\right) \tag{4} \end{eqnarray}
with a similar expression for $I_y$