Is there something called as inrush current at the output of the DC-DC regulator or LDO
As discussed in the comments, your question is somewhat theoretical and not caused by a concrete problem you are experiencing. As such, I'll need to make some assumptions to properly answer the question.
Assuming you are designing a stand-alone device, which plugs into an power source, the proper position to measure inrush current is the source power rail. In your schematic that is the 14V node, so you should put the probe on the rail, before the 220nF capacitor.
This principle does not depend on whether you have an LDO, or a DC-DC regulator, or even if there's no regulator at all. You are measuring whether the power source will cope with the inrush current.
So, on your questions:

Neither pin #1 nor #8 are correct, as these would ignore the charging of the 220nF capacitors, which would partly cause the inrush current.
Also, your suggested probing technique may be inadequate. It really depends on what problems you have with inrush current, but assuming the given schematic is what we're concerned about, the inrush current will be caused by the capacitors on the input and output of the LDO. If the power source can deliver 0.15A (given the limit of the LDO), then the input caps (equiv. to 110nF) charged to 14V will charge in
$$time(s) = \frac{voltage(V) \times capacitance(F)}{current(A)} = \frac{14 \times (110\times10^{-9})}{0.15} = 0.000010267s$$
That's around 10µs. The output caps need around 37µs (using the same formula).
That's assuming the measly 0.15A capability of the power source, and if it's larger then the times decrease proportionately. 10µs is quick, so you definitely want to use an oscilloscope (in a single-shot mode) to measure the current. Then, a current probe around the wire may be inadequate, since such probes are typically bandwidth limited - if the bandwidth is, say, 100 kHz, measuring a 10µs pulse (which is right on the upper end of the bandwidth) would not be accurate as the probe limitations will interfere with the readings.
A better way would be to cut the input wire and place a current shunt resistor (e.g. 1 ohm) and measure the drop across it with an oscilloscope. This way you wouldn't worry much about bandwidth, but the downside is that the addition of the shunt modifies the circuit behaviour a bit, and somewhat reduces the inrush current.
The currents are different, indeed. With my assumptions I've placed already, it is the power source with does not cope with the inrush current drawn from it, so again you should measure the current on the source, same as with point 1.