Note that $\sqrt2\Lambda$ is even, and $\mathrm{vol}(\sqrt2\Lambda)^2=\det(A)$, where $A$ is the Gram matrix of $\sqrt2\Lambda$. Thus $\mathrm{vol}(\Lambda)=2^{-n/2}\sqrt{\det(A)}$, so 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. 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 for a particular $n$ is achieved (usually not uniquely!) by the orthogonal sum of $\Lambda_0$ with some number of 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}
