= Solution
Let $\langle\Phi_{\mathbf k}\Phi_{\mathbf k'}^*\rangle=(2\pi)^3\delta^3(\mathbf k-\mathbf k')P_\Phi(k)$. Statistical isotropy and $\int|Y_{\ell m}|^2d\Omega=1$ applied to the preceding expression give the <CMB angular power spectrum>
$$
C_\ell=\frac{2}{9\pi}\int_0^\infty k^2P_\Phi(k)j_\ell(k\chi_*)^2\,dk.
$$
For $n=1$, define the constant dimensionless potential power $\mathcal A_\Phi=k^3P_\Phi(k)/(2\pi^2)$. Then
$$
C_\ell=\frac{4\pi\mathcal A_\Phi}{9}\int_0^\infty j_\ell(x)^2\frac{dx}{x}.
$$
The needed <logarithmic spherical Bessel square integral> can be proved directly. Put $I_\ell=\int_0^\infty j_\ell^2\,dx/x$ and $J_\ell=\int_0^\infty j_\ell j_{\ell-1}\,dx$. From the <Spherical Bessel function> recurrences,
$$
j_\ell'=j_{\ell-1}-\frac{\ell+1}{x}j_\ell,\qquad
j_{\ell-1}'=\frac{\ell-1}{x}j_{\ell-1}-j_\ell,
$$
integration of the squares' derivatives gives $J_\ell=(\ell+1)I_\ell$, and for $\ell>1$ also $J_\ell=(\ell-1)I_{\ell-1}$. At $\ell=1$, $j_0(0)=1$ and $j_0(\infty)=0$ give $J_1=1/2$, hence $I_1=1/4$. Induction now proves $I_\ell=1/[2\ell(\ell+1)]$. Consequently
$$
\boxed{C_\ell=\frac{2\pi\mathcal A_\Phi}{9\ell(\ell+1)},\qquad
\ell(\ell+1)C_\ell=\frac{2\pi\mathcal A_\Phi}{9}\quad(\ell\geq2).}
$$
This is the flat large-angle <Sachs-Wolfe plateau>. The useful integral printed in the PDF is missing the square on $j_\ell$: without that square the formula is false, since at $\ell=1$ its left-hand side is $\pi/4$, whereas its asserted value would be $1/4$. The squared identity above is the one required by the variance calculation. The plateau describes the matter-era large-angle approximation, not acoustic peaks or late integrated-potential corrections.
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