I did not find a default value of the series resistance in the documentation but printing it gave me 1MOhm.

Good.

As I understand it . . .

. . . wouldn't I still want dt/(cm*R_clamp) = 1 implying R_clamp = dt/cm?

. . . So as I know see it I get the closest estimate if: R_clamp = dt/cm and if I shift the membrane potential and current trace by 1 time step.

No. You're making things more complicated than necessary, and reaching incorrect conclusions, including completely unsupported inferences about what integration time step to choose and mystical ideas about why it is or isn't necessary to shift this or that recording of some current or other by one time step. Throw all that out and start with a blank page.

First, the physics of the situation dictate that in the actual physical system there will always be a lag between the time course of injected current and the resulting change of membrane potential. This is because membrane potential can change only if charge is removed from, or deposited on, membrane capacitance. This must take a finite amount of time unless you are able to apply an infinitely large current (which is impossible).

Second, finite difference numerical integration inherently results in a lag between the time course of injected current and the resulting change of membrane potential in the computational model. To see why, write the differential equation with the derivative on the left, and everything else on the right, so that it is of the form

dv/dt = f(v,t)

To do that, rearrange this

cm*dv/dt + i_ionic = injected_current

so that it becomes

dv/dt = (injected_current - i_ionic)/cm

For a model with linear passive ion channels, this is

dv/dt = (injected_current - g_pas*(v - e_pas))/cm

Let's keep things simple by assuming that the injected current is generated by a current clamp. Then the injected current is independent of v and depends only on time t, and the equation becomes

dv/dt = (i(t) - g_pas*(v(t) - e_pas))/cm

We can simplify this further by assuming that e_pas is 0

Equation 1: dv/dt = (i(t) - g_pas*v(t))/cm

The explicit Euler method approximates the solution to dv/dt = f(v,t) by

v(t+Dt) = v(t) + Dt*f(v(t),t)

where Dt is just a lazy way of writing "delta t".

Applying explicit Euler to Equation 1 yields this recurrence equation

Equation 2: v(t+Dt) = v(t) + (i(t) - g_pas*v(t))*Dt/cm

By inspection it is clear that the injected current at the present moment does not have an immediate effect on membrane potential--i(t) can only affect v at future times.

The implicit, or "backward", Euler method approximates the solution to dv/dt = f(v,t) by

v(t+Dt) = v(t) + f(v(t+Dt),t+Dt)*Dt

Applying implicit Euler to Equation 1, after some rearrangement and grouping of terms, gives us this recurrence equation

Equation 3: v(t+Dt) = v(t)/(1 + Dt*g_pas/cm) + i(t+Dt)*Dt/cm/(1 + Dt*g_pas/cm)

Once again it is clear that the injected current at the present moment does not have an immediate effect on membrane potential--i(t) can only affect v at future times.

So there is always a lag between the injected current and the effect on membrane potential. Furthermore, this is the case even if e_pas is nonzero, and it is still the case if the driving function is current delivered by a voltage clamp that has finite series resistance.

I want to know the current flowing through a certain ion channel inserted in the membrane in response to a given membrane potential ideally resembling the clamped potential.

Easy to do in a computational model. Vector play allows one to force an SEClamp's command potential to follow a piecewise linear approximation to an arbitrary time course. Just make sure that the SEClamp's series resistance is much smaller than the input resistance of the model cell, so that local voltage control is good. As long as the NMODL code that specifies the voltage-gated ion channel's properties exposes the current through those channels to hoc or Python, you're good to go. If it doesn't, that can be easily fixed. If the channel's properties are specified by an instance of the Channel Builder class, nothing special will need to be done.