For a field $\mathbb{F}$, let $\mu_\mathbb{F}(n)$ denote the maximal dimension of a subspace $N\subset A_n(\mathbb{F})$ such that all the nonzero elements of $N$ are invertible. For simplicity, I will assume that the characteristic of $\mathbb{F}$ is not $2$.
Then, because $\det(a) = (-1)^n\det(-a) = (-1)^n\det(a^\mathsf{T})= (-1)^n\det(a)$, we have $\mu_\mathbb{F}(2m{+}1) = 0$.
Meanwhile, clearly, $\mu_\mathbb{F}(2m)\ge 1$, and, as the OP points out, $\mu_\mathbb{R}(4m)\ge 3$ and $\mu_\mathbb{R}(8m)\ge7$ due to the existence of normed division algebras $\mathbb{H}$, of dimension $4$ over $\mathbb{R}$, and $\mathbb{O}$, of dimension $8$ over $\mathbb{R}$.
When $n=2m$, the polynomial function $\det:A_n(\mathbb{F})\to\mathbb{F}$
is the square of a polyomial
$\mathrm{Pf}:A_n(\mathbb{F})\to\mathbb{F}$ homogeneous of degree $m=n/2$, unique up to a choice of sign. In fact, $\mathrm{Pf}$ is defined over the integers, $\mathrm{Pf}:A_n(\mathbb{Z})\to\mathbb{Z}$, as a polynomial with integer coefficients with the property that $\mathrm{Pf}(mam^\mathsf{T}) = \det(m)\,\mathrm{Pf}(a)$ for $a\in A_{2m}(\mathbb{Z})$ and $m\in M_{2m}(Z)$. Consequently, this property holds with $\mathbb{Z}$ replaced by $\mathbb{F}$ for any field $\mathbb{F}$.
It follows that $\mu_\mathbb{R}(4m+2)=1$, since, in this case, $\mathrm{Pf}$ is a polynomial of odd degree, implying that, for any pair $a,b\in A_{4m+2}(\mathbb{R})$, the homogeneous polynomial $p(s,t)=\mathrm{Pf}(sa+tb)$ of odd degree $2m{+}1$ will vanish for some real ratio $[s:t]$.
If $\mathbb{F}$ is an ordered field (more generally, if every nontrivial sum of squares in $\mathbb{F}$ is nonzero), then the standard Clifford algebra construction (using a definite quadratic form) shows that $\mu_\mathbb{F}(n)\ge\rho(n){-}1$, where $\rho(n)$ is the Radon-Hurwitz number. In particular, $\mu_\mathbb{Q}(n)\ge\rho(n){-}1$. Meanwhile, J. F. Adams has shown that $\mu_\mathbb{R}(n)=\rho(n){-}1$. Thus, $\mu_\mathbb{Q}(n)\ge\mu_\mathbb{R}(n)$, but, in general, equality does not hold.
Claim: $\quad 2m{-}1\ge\mu_\mathbb{Q}(2m)\ge m$.
In particular, $\mu_\mathbb{Q}(6)\ge 3 > \mu_\mathbb{R}(6) = 1$, thus verifying that $\mu_\mathbb{Q}(2m)$ can be strictly greater than $\mu_\mathbb{R}(2m)$.
The claim follows from the fact that the characteristic polynomial of a generic element $a\in A_{2m}(\mathbb{Q})$ is irreducible over $\mathbb{Q}$.
For, when the characteristic polynomial of $a$ is irreducible over $\mathbb{Q}$, then $I, a, a^2,\ldots, a^{2m-1}$ spans a field $\mathbb{Q}(a)\subset M_{2m}(\mathbb{Q})$, and hence every nonzero linear combination of these matrices is invertible. Meanwhile, $N(a) = \mathbb{Q}(a)\cap A_{2m}(\mathbb{Q})$ is a vector space with basis $a, a^3, \ldots a^{2m-1}$ and hence has dimension $m$. Thus, $\mu_\mathbb{Q}(2m)\ge m$.
The upper bound follows from the fact that any subspace $N\subset A_{2m}(\mathbb{Q})$ of dimension greater than $2m{-}1$ must intersect nontrivially with the subspace of matrices with the first column equal to zero, since that subspace has codimension $2m{-}1$.
Remark 1: It seems likely that the 'generic' $m$-dimensional subspace of $A_{2m}(Q)$ has all of its nonzero elements invertible, but, perhaps this depends on some carefully defined notion of 'generic'.
Remark 2: Since $\mu_\mathbb{Q}(n)\ge \mu_\mathbb{R}(n)$, the lower bound in the Claim cannot always be strengthened to equality. For example, $\mu_\mathbb{Q}(4)\ge \mu_\mathbb{R}(4) = 3 > 2$. Thus, $\mu_\mathbb{Q}(4)=3$. Similarly, since $\mu_\mathbb{R}(8)=7$, we have $\mu_\mathbb{Q}(8)=7$. (This answers Question 1 for $n=4$ and $n=8$.)
Note that the OP's Question 2 asks whether $\mu_\mathbb{Q}(2m)\ge 2m{-}1$, presumably provoked by the observation that $\mu_\mathbb{R}(2m) = 2m{-}1$, when $m=2$ and $m=4$. However, these low dimensions can be very misleading. For all other values of $m$, we have $\mu_\mathbb{R}(2m) < 2m{-}1$, and, in fact, for all but a finite set of values of $m$, we have $\mu_\mathbb{R}(2m) < m$, and in general, as $m$ grows, the lim inf of $\mu_\mathbb{R}(2m)/m$ equals $0$. On the other hand, $\mu_\mathbb{Q}(2m)/m\ge 1$ for all $m$.
Remark 3: I'm including this last remark at the request of the OP, but, not being a number theorist, I do not have any realy confidence that this can be turned into a rigorous argument.
I do not know whether $\mu_\mathbb{Q}(6)>3$, however, a very heuristic speculation leads me to suspect that this is true and that it might even be true that $\mu_\mathbb{Q}(6)=5$.
The Grassmannian $G_4(15)$ of $4$-dimensional subspaces of $A_\mathbb{Q}(6)$ is a rational variety of dimension $4\cdot (15-4) = 44$. Meanwhile, the group $\mathrm{SL}(6,\mathbb{Q})$ has dimension $35$ and it acts on $A_\mathbb{Q}(6)$ via $m\cdot a = mam^\mathsf{T}$ preserving $\mathrm{Pf}:A_\mathbb{Q}(6)\to\mathbb{Q}$. The induced action of $\mathrm{SL}(6,\mathbb{Q})$ on $G_4(15)$ has generic orbits of dimension $35$, so the 'moduli space' $\mathscr{M}$ of orbits has formal dimension $44-35 = 9$. Meanwhile, the restriction of $\mathrm{Pf}$ to a subspace $N\subset A_\mathbb{Q}(6)$ is a rational cubic form on $N$, generically nondegenerate. The moduli of cubic forms of rank $4$ under $\mathrm{GL}(4,\mathbb{Q})$ equivalence has formal dimension $20 - 16 = 4 < 9$, and it is known that there are rational cubic forms of rank 4 that do not represent $0$ rationally. It seems that the map assigning to a generic $4$-plane $N\subset A_6(\mathbb{Q})$ the rational cubic form $\mathrm{Pf}:N\to\mathbb{Q}$ is 'dominant'. For this reason, it seems likely to me that a 'generic' $4$-plane $N\subset A_6(\mathbb{Q})$ will have the property that $\mathrm{Pf}:N\to\mathbb{Q}$ will not represent $0$ rationally (and hence the nonzero elements of $N$ would all be invertible).
However, it's not that easy to determine whether a given rational cubic forms of rank $4$ represents $0$ rationally, so just choosing a $4$-plane $N$ 'at random' and testing whether its Pfaffian represents $0$ seems to be a very labor intensive way to try to find an example.
All of the above is very speculative, but one could go on to make a similar argument for $5$-planes in $A_6(\mathbb{Q})$. There, it seems even harder to test for when a given rational cubic form of rank $5$ represents $0$ rationally, though.
