Like John Rennie, I'm assuming the circuit is just a resistor and a voltage supply. In that case, you are correct: there should be no current at zero voltage.
The most obvious (to me) source of this "discrepancy" is the uncertainty in your current measurements. (The uncertainties in your voltage measurements is also relevant, but those are probably small enough that you can ignore them.) Your fitted slope and intercept will have uncertainties. Those uncertainties are functions of your raw data points and the uncertainties in those raw data points. I don't know if Excel gives you the uncertainties in your fit parameters; if it does, I wouldn't know how to get it. But you could work that for yourself. Wolfram gives the form of those uncertainties, but I've also seen it in introductory error analysis books. Normally you would want to have your fitting software do that, or write a program to do it. But if Excel won't give you that information, and with only 6 data points, it might not be too difficult to do it by hand. My guess is that you'll find that the uncertainty in the y-intercept is larger than the fitted value you got for the y-intercept, meaning that your data is consistent with a y-intercept of zero.
If you still have access to the equipment, you might try measuring the current with zero applied voltage to see what your multimeter gives you. You might also fill in some more of the region in your plot between $V=0$ and $V=2$. The first would indicate if there is any systematic bias in your multimeter. The second would change your fit parameters and their uncertainties, which might make that discrepancy go away.