A back of the envelope calculation (and that is all this is) would go along the lines of assuming that the white dwarf is made entirely of $^{12}$C (it isn't) and is entirely converted into $^{56}$Ni (it isn't).
The appropriate mass to use would be $\sim 1.4M_{\odot}$ (it is actually a touch lower - the real "Chandrasekhar mass" at which instability sets in is determined by GR collapse; or by inverse beta decay; or by the onset of pyconuclear reactions, all of which take place at $\rho \sim 3 \times 10^{13}$ kg/m$^3$ when the white dwarf has a mass of about 1.37-1.38$M_{\odot}$).
If the star is entirely $^{12}$C, then this means $1.40 \times 10^{56}$ carbon nuclei, containing $1.68\times 10^{57}$ baryons. To conserve the baryon number, the number of $^{56}$Ni nuclei produced is smaller by a factor of 12/56.
The mass of each carbon nucleus (by definition) is $12m_u$, where $m_u$ is the atomic mass unit. The mass of each nickel nucleus is $55.94m_u$.
Thus the change in mass converting all the carbon into nickel is
$$ \Delta M \simeq 1.40\times10^{56}\times 12m_u - 1.40\times10^{56}\times (12/56)*55.94m_u$$
$$\Delta M \simeq 1.8\times 10^{54} m_u = 3.0\times10^{27}\ {\rm kg}$$
Converting this to energy gives $2.7\times 10^{44}$J, which is indeed roughly the energy involved in a type Ia supernova. This is what is responsible for "exploding" the star, since with an initial radius of $\sim 1000$ km, it has a gravitational binding energy, $\sim -3GM^2/5R = -3\times 10^{44}$ J.
A slightly less back of the envelope calculation would include the internal energy of the relativistic electrons, which shrinks the magnitude of the binding energy considerably (it would be exactly zero for a star entirely governed by ideal ultra-relativistic degeneracy pressure and halved for non-relativistic degeneracy pressure), so that a large fraction of the energy released can actually go into photons, neutrinos and the kinetic energy of the ejecta.
Best Answer
The gravitational binding energy is the sum of the gravitational potential energy, $\Omega$, and the total internal kinetic energy, $U$.
If you calculate $\Omega + U$ for a star governed solely by ideal ultra-relativistic electron degeneracy pressure, the net binding energy is zero. This corresponds to the "traditional" Chandrasekhar limit for infinite density and zero radius, which occurs at a mass of $1.44 M_{\odot}$ for a carbon or oxygen white dwarf.
In truth, this situation does not occur in nature. There are a number of small corrections to the equation of state - e.g. electrostatic interactions, but more importantly there are at least two reasons why the white dwarf would become unstable at a mass lower than the canonical Chandrasekhar mass and at finite radius. (i) Neutronisation (aka electron capture) may occur, leading to the removal of degenerate electrons and instability; (ii) If one uses the appropriate Tolman-Oppenheimer-Volkhoff (TOV) general relativistic expression for hydrostatic equilibrium, then the WD becomes unstable (for a carbon WD) at around $1.397M_{\odot}$ and at a small, but finite radius of about 1000 km (see Mathew & Nandy 2017 ).
An approximation of $\Omega \sim -1.5GM^2/R$ (which is valid for a gas governed by relativistic degeneracy pressure - i.e.for a $n=3$ polytrope, http://www.astro.princeton.edu/~gk/A403/polytrop.pdf) gives $\Omega= -6\times 10^{44}$ J.
It is somewhat harder to calculate $U$ on the back of an envelope - you really need to integrate a numerical model in spherical shells, solving the TOV hydrostatic equilibrium equation in GR. However, here goes. Let's get an estimate by using the energy density of gas at the average density of the WD ($6.6\times10^{11}$ kg/m$^3$).
For a carbon gas at this density, the Fermi momentum is $p_F=1.9\times10^{-21}$ kg m/s and the relativity parameter, $p_F/m_e c \simeq 7$. Then, approximating this as ultra-relativistic, the average kinetic energy per electron is $(3/4)p_{F}c$ and the kinetic energy density $u=8.4\times10^{25}$ kg/m$^3$. Multiplying by the stellar volume gives $U=3.5\times10^{44}$ J.
Hence binding energy $\Omega + U = -2.5\times10^{44}$ J.
This is five times the $-5\times 10^{43}$ J quoted in the references you dug out. This could easily be due to my crude approximations in the calculations of $\Omega$ and $U$ (subtracting one big, uncertain number from another), but I also note that in your references they talk about a central density of $2\times10^{12}$ kg/m$^3$, whereas the central density of a WD at the GR Chandrasekhar limit is actually $2.35\times10^{13}$ kg/m$^3$. So I guess their WD is also factor of 2-3 bigger and so their GPE is a factor of 2-3 smaller because of this.
I am puzzled by where this central density comes from (if indeed that's what it is) and would appreciate any comments on this (rather than a downvote).
Footnote:
The OP raises the question of rotation. This might change things. Boshkayev et al. (2013) finds a GR Chandrasekhar limit of 1.386$M_{\odot}$ for non-rotating WDs and a central density of $2.12\times10^{13}$ kg/m$^3$ (consistent with what I use above). The rotating models (shown in Fig.2) show that a WD with 1.38$M_{\odot}$ and central density of 2-3$\times10^{12}$ kg/m$^3$ is possible, but these should be stable - the Chandrasekhar limit is increased by rotation and occurs at lower central densities in these cases, but always I think $\geq 7\times10^{12}$ kg/m$^3$.
Further footnote
After correspondence with one of the authors of the original SN Type 1A progenitor papers, it turns out they are using WD structures that do not use GR in the calculation. Hence the lower central densities at a given mass.