For the purposes of thermodynamics, you can treat the propagating modes of the graviton (and superpartners) as independent scalar fields satisfying
\begin{equation}
\nabla_\mu \nabla^\mu \phi + \xi R \phi = 0.
\end{equation}
And then there's a very quick and dirty way to verify that $AdS_{d + 1}$ will behave like a box. Instead of a minimally coupled scalar with $\xi = 0$, you can consider a conformally coupled scalar with $\xi = \frac{d - 1}{4d}$. This will allow you to perform a Weyl rescaling on the metric to get
\begin{align}
&ds^2 = -\left ( 1 + \frac{r^2}{R^2} \right ) dt^2 + \left ( 1 + \frac{r^2}{R^2} \right )^{-1} dr^2 + r^2 d\Omega_{d - 1}^2 \\
\mapsto \;\; &ds^2 = -dt^2 + \left ( 1 + \frac{r^2}{R^2} \right )^{-2} dr^2 + \left ( 1 + \frac{r^2}{R^2} \right )^{-1} r^2 d\Omega_{d - 1}^2.
\end{align}
We can now get an effective volume of global $AdS_5$ as
\begin{align}
V &= \int \sqrt{g} \, dr \, d\Omega_3 \\
&= \mathrm{Vol}(S^3) \int_0^\infty \left ( 1 + \frac{r^2}{R^2} \right )^{-5/2} r^3 dr \\
&= \frac{2}{3} \mathrm{Vol}(S^3) R^4.
\end{align}
In the present problem there should of course be another piece due to the $S^5$.
To proceed with analyzing a gas of free fields, we know that for a fermion, there can be 0 or 1 quanta with momentum $p$ so we get a single mode partition function of $Z(p) = 1 + e^{-\beta |p|}$. For a boson, there can be arbitrarily many excitations so
\begin{equation}
Z(p) = 1 + e^{-\beta |p|} + e^{-2\beta |p|} + \dots = (1 - e^{-\beta |p|})^{-1}.
\end{equation}
To get the log of the full partition function, we sum over all modes, leading to
\begin{align}
\log Z &= \sum_p \log Z(p) \\
&\approx \int | \log (1 \pm e^{-\beta |p|}) | \frac{V d^d p}{(2\pi)^d} \\
&= \frac{V}{(2\pi \beta)^d} \Gamma(d) \mathrm{Vol}(S^{d - 1}) \zeta^{\pm}(d + 1).
\end{align}
In the last step, we have Taylor expanded the log and integrated term by term. The answer involves the Riemann zeta function $\zeta^-$ for bosons and the so called alternating zeta function $\zeta^+$ for fermions. Knowing that the free energy is
\begin{equation}
F = E - TS = -T \log Z,
\end{equation}
the formula for the entropy now follows from $S = - \frac{\partial F}{\partial T}$. However, the above is overkill if we aren't interested in the precise prefactor.
The slick way to find that $S \sim E^{d/(d + 1)}$ (so $E^{9/10}$ in $9 + 1$ dimensions) is to use scale invariance. Gravitons are massless and cannot introduce a scale beyond the $AdS$ radisu $R$. Therefore at temperatures much larger than $R^{-1}$ (but still much smaller than the string scale), entropy must be extensive. But entropy is a dimensionless quantity so the only way to achieve this is to have
\begin{equation}
S \sim V T^d.
\end{equation}
The same goes for internal energy but that needs to have one more power of temperature
\begin{equation}
E \sim V T^{d + 1}.
\end{equation}
Eliminating $T$ then yields $S$ in terms of $E$. As an aside, a free theory is not the only case where it's possible to determine the prefactor. Another is a 2d conformal field theory where modular invariance leads to the Cardy formula for the density of states.
