To answer this question, a few assumptions must be made:
The Universe may or may not be infinite. It would therefore make sense to answer your question considering the observable Universe only. But the observable Universe increases in size, not only due to expansion (which doesn't add or remove photons), but also because light from more and more distant regions reaches us. In comoving coordinates — i.e. the coordinates that expand with space — the observable Universe has increased in linear size by a factor of 50 since then, so the comoving volume, and hence the total number of photons, has increased by a factor $\gt10^{5}$.
In other words, new photons enter our observable Universe all the time, and they do so at a faster rate than the individual photons lose energy.
Moreover, most of the photons were not created when the CMB was emitted — they had ben around since the end of inflation, scattering aorund on free electrons, until they were "released" at the decoupling/recombination era. I'll address this in the end of my answer.
I know this is not really what you had in mind, so to be specific, I'll compare the total amount of energy in the observable Universe today, to the same space when the CMB was emitted.
Each $\mathrm{cm}^3$ of space holds roughly $n_\mathrm{ph}$ = 411 CMB photons. There are also photons coming from various astrophysical processes (mostly star formation and dust emission), but those are smaller in number by more than two orders of magnitude, and smaller in energy by at least one order of magnitude, probably more (Hill et al 2018), so let's ignore those.
With a CMB temperature of $T_0 = 2.718\mathrm{K}$ (Planck Collaboration et al. 2016), the average energy of a CMB photon is $E_\mathrm{ph,now} = k_\mathrm{B}T_0 = 2.3\times10^{-4}\,\mathrm{eV}$. Since they're seen redshifted by $z\simeq1100$, each photon has lost energy by the same factor. With radius $R = 46.3\,\mathrm{Glyr}$ of the Universe, the total amount of energy lost is
$$
\begin{array}{rcl}
\Delta E & = & E_\mathrm{tot,then} - E_\mathrm{tot,now} \\
& = & (E_\mathrm{ph,then} - E_\mathrm{ph,now}) \,\times\, N_\mathrm{ph,tot}\\
& = & \left((1+z)E_\mathrm{ph,now} - E_\mathrm{ph,now}\right) \,\times\, n_\mathrm{ph} \,\times\, \frac{4\pi}{3}R^3\\
& \simeq & (1+z)E_\mathrm{ph,now} - E_\mathrm{ph,now} \,\times\, n_\mathrm{ph} \,\times\, \frac{4\pi}{3}R^3\\
& \simeq & 4\times10^{88}\,\mathrm{eV}\\
& \simeq & 6\times10^{76}\,\mathrm{erg}\\
& \simeq & 6\times10^{69}\,\mathrm{J},
\end{array}
$$
where the first approximation acknowledges the fact that the energy density today can be neglected compared to the energy density when the photons were emitted.
If you want to know how much energy these photons have lost since they were initially created, at the end of inflation, you just use the redshift corresponding to this epoch — roughly $z\sim10^{26}$. In that case you get some $10^{93}\,\mathrm{J}$.
