Volume of manifolds embedded in $\mathbb{R}^n$

Yes if $N$ is convex - the flow will even increase or decrease distances. No if not convex, looking at $n=2$. The curve length of those parts of $N$ that are concave with respect to $X$ will decrease, and this may balance out the contribution from the convex parts.

I presume by the volume form on $N$ inherited from $\mathbb{R}^n$, you mean the induced hypersurface area measure. I'll write $N_{\epsilon} = \psi_{\epsilon}^X (N)$

The answer to your question is

• Yes if $N$ is mean convex,
• Maybe otherwise depending on $\int_{\partial N} \langle X, \nu \rangle H d\sigma$.

Here's the explanation:

The variation of volume (i.e. hypersurface area) is $$ \partial_{\epsilon}|_{\epsilon = 0} V(N_{\epsilon}) = \int_{\partial N} \langle X, \nu \rangle H d\sigma $$ where $d\sigma$ is the induced measure on the boundary $\partial N$ and $\nu$ is a choice of unit normal vector field and $H$ the mean curvature with respect to $\nu$.

Writing $\eta = \langle X, \nu \rangle$ we see that $\partial_{\epsilon}|_{\epsilon = 0} V(N_{\epsilon}) = 0$ if and only if $$ \int_{\partial N} \eta H d\sigma = 0. $$

The assumption that $X$ is everywhere transverse is that $\eta$ has a sign everywhere. Then in the case that $N$ is mean convex (i.e. $H$ has a sign everywhere) we get $H \eta$ has a sign everywhere and hence $$ \int_{\partial N} \eta H d\sigma \ne 0 $$ so the volume is definitely different for small $\epsilon$.

Even in the case that $N$ is not mean convex, it could be that $\int_{\partial N} \eta H d\sigma \ne 0$ and the volume is different. If $\int_{\partial N} \eta H d\sigma = 0$, then the area is infinitesimally preserved - but it could still be that the volume is different if $\int_{\partial N_{\epsilon}} \eta Hd\sigma \ne 0$ for $\epsilon \ne 0$.