Here is a bit of Mathematica code that rather supports Joseph's conclusion.

Tet[phi_] := {{Sin[phi], 0, Cos[phi]}, {-Sin[phi], 0, Cos[phi]}, {0,
Sin[phi], -Cos[phi]}, {0, -Sin[phi], -Cos[phi]}};

Rect[a_, phi_] := Module[{x, y, u, w},
{x, y, u, w} = Tet[phi];
Polygon[{a x + (1 - a) u, a x + (1 - a) w, a y + (1 - a) w,
a y + (1 - a) u}]];

v = Subsets[Range[4], {3}];

Manipulate[
Graphics3D[{Opacity[0.2], Sphere[{0, 0, 0}, 1], Opacity[0.3],
GraphicsComplex[Tet[phi], Polygon[v]], Opacity[0.8],
Rect[a, phi]}], {phi, 0, Pi/2}, {a, 0, 1}]

Using parameters like a=0.9, phi=1.4 one obtains an elongated rectangle inscribed a flat tetrahedron, close to an equatorial plane. The maximal inscribed ellipse in this rectangle hardly is contained in any triangle that fits in the unit ball.

Edit (2):

The following improved version uses an analytic solution for tetrahedra $ABCD$ with $[AB]$ perpendicular to $[CD]$, and the maximal ellipse $E$ in a plane cutting
$ABCD$ parallel to $[AB]$ and $[CD]$.
In this case the maximal distance of the vertices of the smallest enclosing triangle $T'$ from the origin is

$1 - 2 (1 - a) a (1 + \cos(\phi - \psi))\leq 1$ for $0 < a < 1 $,

where
$A=(\sin \phi, 0, \cos\phi)$, , $B=(-\sin \phi, 0, \cos\phi)$ , $C=(0,
\sin \psi, -\cos\psi)$, $D=(0, - \sin \psi, -\cos\psi)$.
Thus in all these cases the triangle $T'$ lies inside $B$.

I even think that all other (nontrivial) cases of $E\subset T$ can be reduced to one of the above cases by aligning two edges of the enclosing tetrahedron $T$ to lie along the principal axes of $E$, while keeping $E$ inside. But this I cannot prove yet.

Code:

Tet[phi_,
psi_] := {{Sin[phi], 0, Cos[phi]}, {-Sin[phi], 0, Cos[phi]}, {0,
Sin[psi], -Cos[psi]}, {0, -Sin[psi], -Cos[psi]}};
Rect[a_, phi_, psi_] := Module[{x, y, u, w},
{x, y, u, w} = Tet[phi, psi];
Polygon[{a x + (1 - a) u, a x + (1 - a) w, a y + (1 - a) w,
a y + (1 - a) u}]];
v = Subsets[Range[4], {3}];
ParPl[a_, phi_, psi_] :=
ParametricPlot3D[{a Sin[phi] Cos[t], -(-1 + a) Sin[psi] Sin[t],
a Cos[phi] + (-1 + a) Cos[psi]}, {t, 0, 2 Pi}, Boxed -> False,
Axes -> False];
Tangent[a_, phi_, psi_, t_,
lam_] := {a Sin[phi] Cos[t], -(-1 + a) Sin[psi] Sin[t],
a Cos[phi] + (-1 + a) Cos[psi]} +
lam {-a Sin[phi] Sin[t], (1 - a) Cos[t] Sin[psi], 0};

Manipulate[
bt = 2 ArcTan[
1 - (a Csc[psi] Sin[phi])/(-1 + a) - Sqrt[(
a Csc[psi] Sin[phi] (2 - 2 a + a Csc[psi] Sin[phi]))/(-1 + a)^2]];
p1 = Tangent[a, phi, psi, bt, Cot[bt]];
p2 = Tangent[a, phi, psi, bt, -Sec[bt] - Tan[bt]];
p3 = {-1, 1, 1}*p2;
Show[Graphics3D[{Opacity[0.2], Sphere[{0, 0, 0}, 1], Opacity[0.3],
GraphicsComplex[Tet[phi, psi], Polygon[v]], Opacity[0.8],
Rect[a, phi, psi], Green, Polygon[{p1, p2, p3}]}],
ParPl[a, phi, psi]], {{phi, Pi/4}, 0, Pi/2}, {{psi, Pi/4}, 0,
Pi}, {{a, 0.7}, 0, 1}]