What would Mercury's mass have to be in order for it to still have a nitrogen atmosphere?
First find the current $v_{escape}$ and $v_{average molecule}$:

$v_{escape}=\sqrt {\frac{2GM}{r}}= 11.2  km/s \sqrt{\frac{ M  (M_{\oplus})}{R  (R_{\oplus})}}$

(see more precisely 8-1, pg. 208-209)

$M_{Mercury}=3.3 \times 10^{23}  kg =0.055  M_\oplus$
$R_{Mercury}=2400  km= 0.38  R_\oplus$

$v_{escape} (km/s) = 11.2  km/s \sqrt{\frac{ 0.055  (M_{\oplus})}{0.38  (R_{\oplus})}}=4.2 km/s$

So at the surface of the Mercury the escape speed is 4.2 km/s.

The average speed of a nitrogen molecule in Mercury's atmosphere:

$v_{molecule} (km/s)= 0.157 \sqrt{\frac{T_{gas} (K)}{m_{molecule}  (\mbox{units of m$_{H}$})}}$

Use the peak temperature of Mercury: $T_{atmosphere}=700 K$
$m{_{N_2}}=28  (m_H)$

$v_{N_2} (km/s)= 0.157 (km/s)  \sqrt{\frac{ T_{gas}  (K)}{m{_{N_2}} (m_H)}} =0.157 (km/s) \sqrt{\frac{700 (K)}{28 m_{H}}}$

$v_{N_2} (km/s)= 0.8 km/s$

For Nitrogen to remain on Mercury, $6\times v_{average molecule}<v_{escape}$,
but currently $6 \times v_{average N_2}= 4.8  km/s$, which is greater than $v_{escape} (km/s) = 4.2 km/s$, so nitrogen will escape.

What would Mercury's mass have to be in order for it to still have a nitrogen atmosphere?

If we increase the Mass of Mercury, $v_{escape}$ will increase. If
$6\times v_{average N_2}= 4.8  km/s \le v_{escape}$, the nitrogen will remain on Mercury. The condition for Nitrogen to remain is:

$6\times v_{average N_2}= 4.8  km/s \le v_{escape}$

$4.8 km/s \le 11.2  km/s \sqrt{\frac{M}{R}}$

R is fixed, but mass can change. Solve for the new mass:

$\Rightarrow (4.8 km/s)^2 \le (11.2  km/s \sqrt{\frac{M}{R}})^2$

$\left ( \frac{4.8 km/s}{11.2  km/s}\right )^2 \le \frac{M}{R}$

$.18  R \le M$ or $M \ge .18  R$

$M~ (M_{\oplus}) \ge (.18)~ R~(R_\oplus)= (.18) ~ .38~(R_\oplus)$
$M=0.070(M_{\oplus}) = 4.2 \times 10^{23} kg$