What is the direction of the net electric dipole moment in hydronium H3O+?
In aqueous medium, the proton ($\ce{H+}$) is thought to exist as hydronium, $\ce{H3O+}$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be difficult to measure the dipole moment of $\ce{H3O+}$ experimentally. I haven't found any reference that did this, and I suspect it is impossible.
So we have to turn to computational chemistry to calculate the dipole moment of $\ce{H3O+}$. For one isolated $\ce{H3O+}$ (i.e. in gas phase):

The z-axis is set along the $C_3$ axis.
$\text{M06-2X/aug-pcseg-2}$ gives a dipole moment of $\text{1.515 D}$ in z-direction. In x and y direction, the dipole values are negligible:
         DX          DY          DZ         /D/  (DEBYE)
    -0.007834    0.000072    1.515546    1.515566

This means the dipole points along +z-axis in the picture. Partial charges(ESP) fitted to the dipole moment are:
 NET CHARGES:
 -------------------------------------
 ATOM                CHARGE    E.S.D.
 -------------------------------------
 O                  -0.3668    0.0000
 H                   0.4561   -0.0000
 H                   0.4553   -0.0000
 H                   0.4554    0.0000
 -------------------------------------

As you can see, the oxygen atoms has a partial negative charge and hydrogen atoms have partial positive charges. So, the formal charges that we assign to the Lewis structure does not have much significance in this case.
(IUPAC convention of dipole moment is that the arrow points from the negative charge towards the direction of the positive charge. This is exactly opposite of what you wrote in the question)
Another method, $\text{MP2/aug-pcseg-2}$ gives $\text{1.507 D}$ along z-axis, which is close to what we got from the DFT calculation.
(Note that I have not enforced symmetry for the calculation, which is why the partial charges are slightly different for each hydrogen. This is a numerical error, because in reality all the hydrogens are equivalent)