The set of the values of $\{n/\{\sqrt 2 n\}\}$ is dense in $[0,1)$.
The choice of $n$ is as follows: $n=ka_p+1$, where $a_p$ is the $p$th term of A001541. For fixed $k \in \mathbb N^+$, we let $a_p\rightarrow \infty$, and the value of $\{n/\{\sqrt 2 n\}\}$ tends to $\{-\frac {\sqrt{2}} 4 (3k^2-4)\}$$\{-\frac {\sqrt{2}} 2 (3k^2-2)\}$.
Proof: Let $b_p=\sqrt{2a_p^2-2}$. By the properties of A001541, $b_p$ are all integers, and $b_p-\sqrt2 a_p=O(1/a_p)$.
Thus, for fixed $k$ and sufficiently large $p$, $\{\sqrt2 n\}=\{\sqrt2 ka_p+\sqrt{2}\}=\sqrt{2}ka_p-kb_p+\sqrt2-1$.
Since $\sqrt{2}ka_p-kb_p$ is $O(1/a_p)$, we may compare $n/\{\sqrt 2 n\}$ to $n/(\sqrt 2 - 1)$:
$n/\{\sqrt 2 n\}=\frac{ka_p+1}{\sqrt{2}ka_p-kb_p+\sqrt2-1}=\frac{ka_p+1}{\sqrt2-1}-\frac{(\sqrt{2}ka_p-kb_p)(ka_p+1)}{(\sqrt2-1)(\sqrt{2}ka_p-kb_p+\sqrt2-1)}$.
The first term equals to $(\sqrt2+1)ka_p+(\sqrt2+1)$, which is mod-1 equal to $\sqrt2ka_p-kb_p+\sqrt{2}$. It mod-1 converges to $\sqrt2$. So we only need to prove that the second term, without the minus sign, mod-1 converges to $\frac{3\sqrt2}2k^2$.
To see this, notice that the leading term of the second term is $ \frac{(\sqrt{2}ka_p-kb_p)ka_p}{(\sqrt2-1)^2}$. As $b_p^2=2a_p^2-2$, the term can be rewritten as $k^2\frac{2a_p}{\sqrt2a_p+b_p}(3+2\sqrt2)$. The term tends to $k^2\frac{\sqrt{2}}{2}(3+2\sqrt2)$, which is mod-1 equivalent to $\frac{3\sqrt2}2k^2$.
By the theorem of Weyl as you mentioned, when $k$ runs over $\mathbb N^+$, the set $\{-\frac {\sqrt{2}} 4 (3k^2-4)\}$$\{-\frac {\sqrt{2}} 2 (3k^2-2)\}$ is dense in $[0,1)$.
Thus the set of the values of $\{n/\{\sqrt 2 n\}\}$ is dense in $[0,1)$.