The height of the mercury in the barometer would be less.

$P = \rho gh$, without acceleration

$h = \frac{P}{\rho g}$, without acceleration

$P = \rho (g + a)h$ with acceleration

$h = \frac{P}{\rho (g +a)}$ with acceleration

The actual atmospheric presure would be the same in either case, but the barometer would report a lower pressure in units of cm or mm of mercury.

If the elevator and pool of mercury was exposed instead of a normal enclosed elevator, then movement relative to the air would create an additional pressure. The pressure from wind is very small at the speed of an elevator, and can be calculated by:

$P_{wind} = \frac{1}{2}\rho_{air} v^2$, where $v$ is the velocity of the wind (elevator in this case).

$\rho_{air} =1.25 kg/m^3$

The height of the mercury in the barometer would be less.

$P = \rho gh$, without acceleration

$h = \frac{P}{\rho g}$, without acceleration

$P = \rho (g + a)h$ with acceleration

$h = \frac{P}{\rho (g +a)}$ with acceleration

The actual atmospheric presure would be the same in either case, but the barometer would report a lower pressure in units of cm or mm of mercury.

If the elevator and pool of mercury was exposed instead of a normal enclosed elevator, then movement relative to the air would create an additional pressure. The pressure from wind is very small at the speed of an elevator, and can be calculated by:

$P_{wind} = \frac{1}{2}\rho_{air} v^2$, where $v$ is the velocity of the wind (elevator in this case).

$\rho_{air} =1.25 kg/m^3$

So at a speed of $1 m/s$ , the additional pressure is only $0.6Pa$ compared to atmospheric pressure of $101325 Pa$!