One more exact answer: for $n=8$ the minimal dimension is $22$, again attaining the "easy" bound ${n-1 \choose 2} + 1$ and the same as the value for the next dimension $n=9$.

First to explain why ${n-1 \choose 2} + 1$ is enough, not just for real vector spaces but for an $n$-dimensional vector space $V$ over any field. Fix nonzero $X_0 \in V$. Then $X_0 \wedge V = \{ X_0 \wedge Y : Y \in V \}$ is a linear subspace of dimension $n-1$ in $\bigwedge^2 V$. Thus if $E \subset \bigwedge^2 V$ is a linear subspace of codimension $n-2$ then it must have nonzero intersection with $X_0 \wedge V$.

Now take $n=8$ and identify $V$ with the Cayley octonions. Let $V_0 \subset V$ consist of the "purely imaginary" octonions, so $V$ is the orthogonal direct sum of $\bf R$ with $V_0$. Write $\bigwedge^2 V = ({\bf R} \wedge V_0) \oplus \bigwedge^2 V_0$, and let $E$ be the kernel of the homomorphism $h: \bigwedge^2 V \to V_0$ that takes $1 \wedge X$ to $X$ and $X \wedge Y$ to the imaginary part of the octonion $XY$, for any $X,Y \in V_0$. [This is well-defined because $XY + YX \in \bf R$ for all $X,Y \in V_0$, so $h(Y\wedge X) = - h(X \wedge Y)$.] Then $E$ has dimension $21$. I claim that $E$ contains no nonzero pure tensors. Indeed a pure tensor in $\bigwedge^2 V$ has the form $1 \wedge X$, $X \wedge Y$, or $(1+X) \wedge Y$ for some $X,Y \in V_0$ which are linearly independent (so in particular nonzero). Certainly $h(1\wedge X) = X$ is nonzero. So is $h(X \wedge Y)$, because if $XY \in \bf R$ for $X,Y \in V_0$ then $X$ and $Y$ are proportional. Finally $h\bigl((1+X) \wedge Y\bigr) = X + h(X \wedge Y)$ cannot be zero because the imaginary part of $XY$ is orthogonal to $X$.

It also follows that for $n=6,7$ the minimal $\dim E$ is at least ${n \choose 2} - 6 = 9, 15$, and thus exceeds the lower bound ${n-2 \choose 2} + 1 = 7, 11$ coming from the dimension of the Grassmannian.

Replacing the Cayley octonions by the Hamilton quaternions recovers the answer of $4$ for $n=4$.

One more exact answer: for $n=8$ the minimal dimension is $22$, again attaining the "easy" bound ${n-1 \choose 2} + 1$ and the same as the value for the next dimension $n=9$.

First to explain why ${n-1 \choose 2} + 1$ is enough, not just for real vector spaces but for an $n$-dimensional vector space $V$ over any field. Fix nonzero $X_0 \in V$. Then $X_0 \wedge V = \{ X_0 \wedge Y : Y \in V \}$ is a linear subspace of dimension $n-1$ in $\bigwedge^2 V$. Thus if $E \subset \bigwedge^2 V$ is a linear subspace of codimension $n-2$ then it must have nonzero intersection with $X_0 \wedge V$.

Now take $n=8$ and identify $V$ with the Cayley octonions. Let $V_0 \subset V$ consist of the "purely imaginary" octonions, so $V$ is the orthogonal direct sum of $\bf R$ with $V_0$. Write $\bigwedge^2 V = ({\bf R} \wedge V_0) \oplus \bigwedge^2 V_0$, and let $E$ be the kernel of the homomorphism $h: \bigwedge^2 V \to V_0$ that takes $1 \wedge X$ to $X$ and $X \wedge Y$ to the imaginary part of the octonion $XY$, for any $X,Y \in V_0$. [This is well-defined because $XY + Y\!X \in \bf R$ for all $X,Y \in V_0$, so $h(Y\wedge X) = - h(X \wedge Y)$.] Then $E$ has dimension $21$. I claim that $E$ contains no nonzero pure tensors. Indeed a pure tensor in $\bigwedge^2 V$ has the form $1 \wedge X$, $X \wedge Y$, or $(1+X) \wedge Y$ for some $X,Y \in V_0$ which are linearly independent (so in particular nonzero). Certainly $h(1\wedge X) = X$ is nonzero. So is $h(X \wedge Y)$, because if $XY \in \bf R$ for $X,Y \in V_0$ then $X$ and $Y$ are proportional. Finally $h\bigl((1+X) \wedge Y\bigr) = Y + h(X \wedge Y)$ cannot be zero because the imaginary part of $XY$ is orthogonal to $Y$.

It also follows that for $n=6,7$ the minimal $\dim E$ is at least ${n \choose 2} - 6 = 9, 15$, and thus exceeds the lower bound ${n-2 \choose 2} + 1 = 7, 11$ coming from the dimension of the Grassmannian.

Replacing the Cayley octonions by the Hamilton quaternions recovers the answer of $4$ for $n=4$.