Arduino auto power of circuit possible issue with higher voltage agains output pin?
No.  It is not possible for a modern IC such as an MCU having I/O protection diodes to enable/disable its own power using a single external N-Channel or P-Channel device.  The reason is that the path broken by the single switch ends up being partially completed through I/O pin's internal protection diode and your pulldown resistor, and depending on the voltage drop that results across that, partial turn on the FET itself.  This is enough to have the processor at least partially on - perhaps not enough to operate, but enough to keep consuming battery, and in some cases also enough to make clean startup on manual operation of the power switch unworkable.

To make this work, you need two stages of switching - two fets are a possibility.

However another easy one is to use one of the USB power switches chips intended to be included in the rarely seen "full" USB hubs that have software control of the downstream ports' power.  These are hardly ever fitted in the hubs sold, but they exist as a part, are easy to apply, typically have low leakage, and are often specified to operate correctly both substantially below and somewhat above the 5v they would see in a USB application.  Most importantly, they are multi-stage solutions in a single package - typically they function as high side switches, but they are enabled by an input going high, and as result, they do not suffer the parasitic partial turn on issue via the MCU's protection diodes in the way that a high side switch enabled by an input going low would.  I believe the RT9701 that I used in the past to do this with an ESP8266 is officially discontinued, but still available and likely replaced by newer similar parts.

Given your input voltage, another thing to consider would be using a regulator with an active-high enable input.  But realize that a linear regulator is an inherently wasteful device, and a lot of the compact switchers don't have very high input voltage allowance.