Note that $\sqrt2\Lambda$ is even, and $\mathrm{vol}(\sqrt2\Lambda)^2=\det(A)$, where $A$ is the Gram matrix of $\sqrt2\Lambda$. Thus $\mathrm{vol}(\Lambda)=2^{n/2}\sqrt{\det(A)}$. Thus the smallest possible volume in dimension $n$ is determined by the smallest determinant of an even $n$-dimensional lattice.

The smallest determinant question is answered in the lemma of section 5 in Conway & Sloane's [Lattices with Few Distances](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.391.9846&rep=rep1&type=pdf). The answer is $8$-periodic in $n$, given by

| $n\ (\mathrm{mod}\ 8)$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|-:|-:|-:|-:|-:|-:|-:|-:|-:|-:|
|$\Lambda_0$ | $\{0\}$  | $A_1$ | $A_2$ | $D_3$ | $D_4$ | $D_5$ | $E_6$ | $E_7$ | $E_8$ |
| $\det A$  | $1$ | $2$ | $3$ | $4$ | $4$ | $4$ | $3$ | $2$ | $1$ |

where the minimum is achieved (in general not uniquely!) by the orthogonal sum of $\Lambda_0$ with several copies of $E_8$.

That one cannot do better follows from the classification of even forms of small determinant, SPLAG 15.8, table 15.4: if $\det A<4$ then the mod-8 signature is $\pm (\det A-1)$.

Returning to the original question, this shows
\begin{align}
\liminf_{n\to\infty}\ 2^{n/2}\mathrm{vol}(\Lambda)	=1,\\
\limsup_{n\to\infty}\ 2^{n/2}\mathrm{vol}(\Lambda)	=2.
\end{align}