There is no such set $S$ for even $n\ge4$. We first handle the case $n=4$ by a parity argument: Below we provide a set $V\subset\mathbb R^4$ of $18$ distinct vectors and a set $T$ of $9$ orthonormal bases (with vectors from $V$) such that each vector from $V$ is contained in exactly $2$ bases from $T$.

Now double count the set
\begin{equation}
M=\{(v, t)\;|\;v\in V\cap S, v\in t, t\in T\}.
\end{equation}
Given $v\in V\cap S$, there are exactly two $t\in T$ such that $(v,t)\in M$. So $\lvert M\rvert$ is even. On the other hand, given one of the $9$ possibilities of $t$, the number of $v$'s such that $(v,t)\in M$ is odd by assumption. Thus $\lvert M\rvert$ is odd, a contradiction.

For simplicity of notation, in the following example the vectors differ from normed vectors by a positive factor. As a verification by hand might be a little cumbersome, I provide the straightforward Python code which proves the assertion:

```
from itertools import combinations, chain
T = [[(0, 1, 0, 0), (0, 0, 1, -1), (1, 0, 0, 0), (0, 0, 1, 1)],
[(0, 0, 1, -1), (1, 1, 1, 1), (1, 1, -1, -1), (1, -1, 0, 0)],
[(0, 1, 0, 1), (0, 0, 1, 0), (1, 0, 0, 0), (0, 1, 0, -1)],
[(0, 1, 0, 0), (1, 0, 0, -1), (0, 0, 1, 0), (1, 0, 0, 1)],
[(1, 1, -1, 1), (1, -1, 0, 0), (0, 0, 1, 1), (1, 1, 1, -1)],
[(1, -1, 1, -1), (1, 0, -1, 0), (1, 1, 1, 1), (0, 1, 0, -1)],
[(0, 1, 0, 1), (1, 1, 1, -1), (1, -1, 1, 1), (1, 0, -1, 0)],
[(1, -1, 1, -1), (1, 0, 0, 1), (0, 1, 1, 0), (1, 1, -1, -1)],
[(1, 1, -1, 1), (1, 0, 0, -1), (0, 1, 1, 0), (1, -1, 1, 1)]]
V = list(chain(*T))
def inner_prod(u, v):
return sum(x*y for x, y in zip(u, v))
def is_ortho(t):
return all(inner_prod(u, v) == 0 for u, v in combinations(t, 2))
assert all(is_ortho(t) for t in T)
assert all(V.count(v) == 2 for v in V)
```

Next we show that if $\mathbb R^n$ does not have such a set $S$, then neither does $\mathbb R^{n+2}$. The argument is inspired by Alex Ravsky's answer, but I believe it is easier to digest:

In order to achieve a contradiction, we assume that $\mathbb R^{n+2}$ has such a set $S$. Let $e_1=(1,0,\ldots),\ldots,e_{n+2}=(\ldots,0,1)$ be the standard basis vectors of $\mathbb R^{n+2}$. As $n+2\ge3$, there are two of these vectors which either are both in $S$ or none of them is in $S$. By reordering the coordinates, we may and do assume that this applies to $e_{n+1}$ and $e_{n+2}$.

For $v\in\mathbb R^n$ let $\hat v\in\mathbb R^{n+2}$ be $v$ extended by $(0,0)$. Set
\begin{equation}
S'=\{v\in\mathbb R^n\;|\;\hat v\in S\}.
\end{equation}
Now let $v_1,\ldots,v_n$ be an arbitrary orthonormal basis of $\mathbb R^n$. Then $\hat v_1,\ldots,\hat v_n,e_{n+1},e_{n+2}$ is an orthonormal basis of $\mathbb R^{n+2}$. By assumption, an odd number of them is contained in $S$, and this number equals modulo $2$ the number of $v_i$'s in $S'$. So $S'$ is an admissible set for $\mathbb R^n$, a contradiction.

*Remark:* I do not know if there is a more conceptual description of the given or similar examples for $n=4$. This example has some symmetries: Consider the graph whose vertex set is $V$ and two vertices are connected if and only if their vectors are orthogonal. Then this graph is vertex transitive with an automorphism group of order $72$. Attempts with smaller vertex transitive graphs were not successful.