You answer is indeed correct, and this is the foundation of "raytracing", a widely-used technique in graphics. Typically in graphics we are not given $T$ but rather its inverse (the "modeling transform"), so the formulas you'll see in graphics book will have a $T^{-1}$ in them ... but that's because they're talking about a slightly different problem.
If you want to generalize this to a more complicated transformation, $T$, that might include both translation and a linear part, there's a subtlety: in transforming the ray, you need to transform the start-point by $T$, but transform the direction vector by $DT$, the derivative of $T$. When $T$ is linear, the derivative is just $T$, so your formula works fine. When it's got a translation added, the derivative exactly deletes the translation, and the formula needs to take this into account.
(If you also want to transform the normal vector to the sphere, you need to transform that by the transpose of the derivative of $T$, more or less; it depends on how you intend to use the normal vector. (In particular, the transformation I mentioned may change its length, which is a problem if you need a unit vector!)
General approach of computing a gradient
Your axis-aligned ellipsoid can be rewritten as
$$ f(x,y,z) := b^2c^2x^2 + a^2c^2y^2 + a^2b^2z^2 = a^2b^2c^2 $$
Now the gradient of that left hand side consists of the partial derivatives.
$$ \vec\nabla f = \begin{pmatrix}
2b^2c^2x \\
2a^2c^2y \\
2a^2b^2z
\end{pmatrix} $$
Sinde for the normal direction the magniture is irrelevant, you might drop that factor $2$ in each of these terms.
For a different orientation, make sure to write the ellipsoid as a polynomial in $x,y,z$. Then you can apply the same technique of computing partial differentials.
Finding the formula for rotated spheroid
If you take the spheroid $\frac{x^2+y^2}{a^2}+\frac{z^2}{c^2}=1$ and rotate it so that the original $z$ axis aligns with a vector $r=(s,t,u)$, what equation do you get? The OP actually asked this very question, and in my answer there I came up with the following equation for the spheroid:
\begin{multline*}
c^2\bigl((tz-uy)^2+(ux-sz)^2+(sy-tx)^2\bigr) + a^2(sx+ty+uz)^2 \\
= a^2c^2(s^2+t^2+u^2)
\end{multline*}
Now expand that, collect terms with common monomials, and do the partial derivatives as above. I used a bit of computer algebra here.
$$\vec\nabla f = 2\begin{pmatrix}
a^2s^2+c^2(t^2+u^2)&(a^2-c^2)st&(a^2-c^2)su\\
(a^2-c^2)st&a^2t^2+c^2(s^2+u^2)&(a^2-c^2)tu\\
(a^2-c^2)su&(a^2-c^2)tu&a^2u^2+c^2(s^2+t^2)
\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}$$
Best Answer
Use the inverse transpose of T.
E.g. http://explodedbrain.livejournal.com/112906.html
Or https://www.scratchapixel.com/lessons/mathematics-physics-for-computer-graphics/geometry/transforming-normals.