Some of the comments here seem to be suggesting that there should not be any residual charge of the Sun at all because of the fact that in a conducting medium no electric fields can exist. This argument ignores the crucial point here, namely that there are unequal numbers of positive and negative charges, because electrons, unlike ions, can easily escape from the gravitational field of the sun (in fact, virtually all of them would escape without an electric field holding them back). And any object with an unequal number of positive and negative charges will appear charged from the outside.
The net charge resulting from the escape can easily be calculated from the fact that any particle with a kinetic energy higher than the absolute value of the combined potential energy due to gravity and any net charge $Q$ inside a sphere of radius $R$ will escape from the Sun. So for an electron this applies for energies
$$K_e > \phi_e =\frac{GMm_e+Qe}{R}$$
and for ions for energies
$$K_I > \phi_I = \frac{GMm_I-Qe}{R}$$
where $G$ is the gravitational constant, $M$ the mass of the sun and $e$ the absolute value of the elementary charge (using cgs-units here)
In order to achieve a kind of steady state, we must have the same amounts of positive and negative charges escaping, i.e. we must have for the energy distribution functions
$$f_e(K/\phi_e)=f_I(K/\phi_I)$$
where $K$ is now taken as a general energy variable.
In thermal equilibrium, the distribution functions of the electrons and ion will be the same i.e. $f_e=f_I=f$ (it should be given by the Maxwell-Boltzmann distribution, but knowing the exact form is not even required here), which means we have the condition $\phi_e=\phi_I$ (in other words, for electrons and ions of the same kinetic energy to have the same escape rates, they must have the same potential energy) i.e.
$$GMm_e+Qe = GMm_I-Qe$$
and thus
$$Q=\frac{GM}{2e}(m_I-m_e)$$
Inserting the constants for this (with $m_I$ the proton mass) and converting to SI units gives $Q=77$ Coulombs for a star the mass of the sun (this value is identical to the one derived in the paper by Neslusan (which has already been mentioned a couple of times on SE), but I think my derivation here is more straightforward and easier to understand).
It is remarkable that the charge does only depend on the mass of the star and not for instance on the plasma energy.
For the electric field near the surface of the sun at radius $R$ we get therefore from Coulomb's law
$$E=\frac{Q}{R^2} = 1.4\times 10^{-6} \frac{V}{m}$$
(after again converting from cgs to SI units).
This electric field is very small. It means that over the size of an atomic orbit the corresponding electric potential energy varies only by about $10^{-16} eV$. This would change the wavelength of spectral lines only by an amount which is 12 orders of magnitude smaller than the observed width of spectral lines, so spectroscopically this is impossible to detect.
However, as you mentioned the solar wind: the fact that this is observed to be quasi-neutral to a high degree shows trivially that the sun must be positively charged by the amount derived above. If the sun would be perfectly neutral there would be a massive excess of electrons in the solar wind (of course, this would then in turn charge up the sun, so such an assumption would be logically inconsistent in the first place).
It should also be of observable relevance for modelling the solar atmosphere, because the electric field, although very small, effectively halves the gravitational acceleration for ionized atoms, hence resulting in twice the density scale height compared to the neutral atmosphere (a fact that is also well known from observations of the earth's ionosphere).
As far as direct direct experimental verification is concerned, one should not overlook the fact that the electrostatic force on an ion is not only $-1/2\times$ the gravitational force near the sun (as follows from the above theoretical consideration) but in principle also at any other distance. At the earth, both should be about a factor $2\times 10^{-5}$ smaller, so the electric field would lead to an acceleration of an ion of about $3\times 10^{-4}$ the earth's gravitational acceleration of $9.81 m/s$. Within a couple of minutes, an ion initially at rest would therefore be accelerated to a speed of the order of $1m/s$ due to the Sun's charge. The problem will obviously be to eliminate any other electric fields whilst avoiding shielding the field by the experimental setup. I don't know whether this is technically feasible in practice, but in principle it should be possible. Gravimeters are considerably more sensitive than this these days, so at least the effect of both the Earth's and the Sun's gravity could easily be subtracted from the observed acceleration.
