Given your initial answer for $\mathbf R^2,$ to move to $\mathbf R^3$ consider the vectors $(x,y,z)$ such that $x,y,z \geq 0$ that are equiangular between the plane vectors $(1,0,0)$ and $(1,1,0).$ These make up the plane
$$ y = (\sqrt 2 - 1) x $$ with arbitrary $z \geq 0.$ In order to get the same angle with
$(1,0,1)$ we get the ray
$$ (\sqrt 2 - 1) x = y = z. $$ In order to get the same angle with
$(1,1,1)$ we get the ray
$$ y = (\sqrt 2 - 1) x, \; \; z = (\sqrt 3 - \sqrt 2) x. $$
Note that $ (\sqrt 2 - 1) = 0.4142... $ while $ (\sqrt 3 - \sqrt 2) = 0.317837...$ So in $\mathbf R^3$ the latter comes first while increasing $z,$, and best vector is $(1, \; \;\sqrt 2 - 1, \; \; \sqrt 3 - \sqrt 2) $ where you can work out the angle.

The same process gets you from $\mathbf R^3$ to $\mathbf R^4,$ take this answer for $\mathbf R^3$ and increase the fourth coordinate until you have an equal angle with $(1,1,1,1).$

And so on.

EDIT: I get it. In $\mathbf R^n$ the optimal vector is
$$ V = (1, \; \;\sqrt 2 - 1, \; \; \sqrt 3 - \sqrt 2, \ldots, \sqrt n - \sqrt {n-1})$$ with equal angles to
$A_1 = (1,0,\ldots,0),$ $A_2 = (1,1,0,\ldots,0),$ $A_3 = (1,1,1,0,\ldots,0), \ldots,$ $A_n = (1,1,1,\ldots,1).$

EDIT 2: Note that the entries of $V$ are strictly decreasing. As a result, if instead we consider $B_k$ with $k$ entries set to $1$ and the other $n-k$ set to $0,$ then
$$ | A_k | \; = \; | B_k | $$
but
$$ V \cdot A_k > V \cdot B_k .$$ Therefore the angle between $V$ and $B_k$ is larger than the angle between $V$ and $A_k,$ and the angle we actually constructed is the best we can get away with.