It is called the de Sitter temperature and is given by $T = H/2\pi$, where $H$ is the Hubble parameter. It arises from the autocorrelation of superhorizon quantum fluctuations generated during inflation.
EDIT: Below, I elaborate on the above showing how the autocorrelation of field fluctuations during inflation leads to a de Sitter temperature $T=H/2\pi$.
Begin with the inflaton, $\phi$, a minimally-coupled scalar field evolving in a FRW universe:
$$\ddot{\phi} + 3H \dot{\phi} – \frac{\nabla^2}{a^2}\phi + \frac{{\rm d}V(\phi)}{{\rm d}\phi} = 0$$
and consider small perturbations about its homogeneous background value, $\phi({\bf x},t) = \phi_0(t) + \delta \phi({\bf x},t)$. To first order in $\delta \phi({\bf x},t)$, the above expression becomes
$$\ddot{\delta \phi} + 3H\dot{\delta \phi} -\left(\frac{\nabla^2}{a^2} – \left.\frac{{\rm d}^2V(\phi)}{{\rm d}\phi^2}\right|_{\phi = \phi_0}\right)\delta \phi = 0.$$
Next, take the Fourier transform of the fluctuation in comoving wavenumber, $k$,
$$
\delta \phi({\bf x},t) = \int\frac{{\rm d}^3k}{(2\pi)^{3/2}}\delta \phi_k(t)e^{i{\bf k}\cdot{\bf r}},
$$
and use it to obtain an expression for the Fourier modes $\delta \phi_k$,
$$\ddot{\delta \phi_k} + 3H\dot{\delta \phi_k} + \left(\frac{k}{a}\right)^2\delta \phi_k = 0.$$
Lastly, we can rescale the field $u_k = a\delta \phi_k$ to arrive at the more compact mode equation,
$$ u_k'' + \left(k^2 – \frac{a''}{a}\right)u_k = 0,
$$
where primes denote derivative wrt conformal time, ${\rm d} \tau = {\rm d}t/a$.
Specializing to the case of de Sitter space for which $a(t) \propto e^{Ht}$ and $\tau = -1/aH$ we can write the mode equation as
$$(k\tau)^2\frac{{\rm d}^2u_k}{{\rm d}(k\tau)^2} + \left[(k\tau)^2 – 2\right]u_k = 0.$$
This can be solved exactly in terms of Hankel functions,
$$u_k(-k\tau) = \frac{1}{2}\sqrt{-k \tau}\left[c_1 H^{(1)}_{3/2}(-k\tau) + c_2H^{(2)}_{3/2}(-k\tau)\right],$$
where $H^{(1)}_{3/2} = J_{3/2} +iY_{3/2} = H^{(2)*}_{3/2}$ and $J$ and $Y$ are Bessel functions of the first and second kind.
Now, let's figure out the constants. We can get one of them by appealing to a generic feature of quantum fields in expanding spacetimes: when the frequency is high relative to the expansion rate, the field doesn't "feel" the expansion and it oscillates as a plane wave. In the short wavelength limit $k/aH = -k\tau \rightarrow \infty$, Hankel functions indeed reduce to sinusoids,
$$u_k(-k\tau) = \frac{1}{\sqrt{2k}}\left(c_1 e^{-ik\tau} + c_2 e^{ik\tau}\right).$$
To recover positive frequency plane waves, we choose $c_2 = 0$.
To figure out $c_1$, we make use of the fact that $\delta \phi({\bf x},t)$ is a quantum mode, and so it must satisfy the so-called canonical commutation relation with its conjugate momentum, $\pi({\bf x},t) = a^2\delta \phi({\bf x},t)’$. This is a straight-forward but tedious calculation involving a small collection of Bessel function identities: you should find at the end that $c_1 = \sqrt{\pi/k}$.
The final expression governing the time-dependence of a quantum mode in de Sitter space can now be written down,
$$
u_k(-k\tau) =-\frac{\sqrt{\pi \tau}}{2}H^{(1)}_{3/2}(-k\tau)=-\frac{1}{\sqrt{2k}}\left(1 – \frac{i}{k\tau}\right)e^{-ik\tau}.$$
This is a wonderful result, particularly the second equality which results because Bessel functions of half-integer order are just combinations of trig functions. It's incredibly insightful: the fluctuation starts as a plane wave in the distant past when its wavelength is tiny ($-k\tau \rightarrow \infty$), but then as the mode is stretched by the expansion ($-k\tau \rightarrow 0$), it evolves out of the vacuum ultimately obtaining a constant amplitude on large scales (see figure). This is called mode freezing, and is due to the quantum decoherence of fluctuations on super-horizon scales.
Specifically, in the large wavelength limit we can use small-argument approximations of the Bessel functions to see that $H^{(1)}_{3/2}(-k\tau)$ becomes the constant $\sqrt{2/\pi}(k\tau)^{-3/2}$ and
$$
|\delta \phi_k| = \frac{|u_k|}{a} = \frac{H}{\sqrt{2k^3}}.$$
This is the quantity of interest, since it gives the amplitude of the fluctuation on large scales where it is a real, classical perturbation.
The correlation function is then obtained by ensemble averaging,
$$
\langle | \delta \phi_k|^2\rangle = \frac{H^4}{2k^3}.
$$
To get a quantity independent of scale, it is common practice to multiply by $k^3/2\pi^2$ to get the power spectrum,
$$
P(k) = \frac{H^2}{4\pi^2}.
$$
This has dimensions of energy-squared, and so its square root corresponds to a temperature, the Gibbons-Hawking temperature, of $T = H/2\pi$.
