Here's a very quick-and-dirty estimate. Let's assume the earth is a sphere of radius $R$, that we're at latitude $\phi$, that the sun is a point source, and sets due west, in the plane containing the latitude circle of our location, i.e., it's an equinox.
Looking down from the nearest pole, we see our latitude as a circle of radius $R\cos\phi$. At sunset, we see the sun's ray as the line tangent to the latitude at our location. In polar coordinates $(r, \theta)$ with origin at the center of our latitude, this line has equation $r = R\cos\phi \sec\theta$. The shadow's distance from the latitude circle at (longitude) angle $\theta$ (in radians) is therefore
$$
r - R\cos\phi = R\cos\phi(\sec\theta - 1)
\approx \tfrac{1}{2}R\cos\phi \cdot \theta^{2}.
$$
(In the diagram, our location is the rightmost point on the latitude circle and the vertical line represents the sun's ray through our location, shining upward. The approximation is the second-order Taylor approximation of secant.)
Edit: As David K notes in the comments, the preceding direction makes angle $\frac{\pi}{2} - \phi$ with the surface of the earth. A more accurate estimate of the altitude of the shadow above the ground is therefore obtained if we multiply by $\cos\phi$:
$$
\text{Altitude} \approx \tfrac{1}{2}R\cos^{2}\phi \cdot \theta^{2}.
$$
(This formula agrees to second order with the estimate obtained by modeling the earth's shadow as a cylinder circumscribed on the earth with the sun lying on the axis at infinite distance; calculation omitted.)
The earth rotates one full turn in $86400$ seconds, so $t$ seconds after sunset the earth has rotated by an angle
$$
\theta = \frac{2\pi t}{86400},
$$
and the altitude of the shadow (adjusted as noted above) has increased to
$$
\tfrac{1}{2}R\cos^{2}\phi \cdot \theta^{2}
\approx 3175 \cos^{2}\phi \frac{(2\pi)^{2}}{86400^{2}} t^{2}
\approx 1.68 \cos^{2}\phi \times 10^{-5} t^{2}
$$
kilometers. At latitude $32.5^{\circ}$ this becomes about $1.2 \times 10^{-5} t^{2}$. Thus:
$$
\begin{array}{l|lllll}
\text{Minutes after sunset at sea level} & 1 & 2 & 3 & 4 & 5 \\
\text{Sun is setting at altitude (in m)} & 43 & 172 & 387 & 688 & 1075 \\
\end{array}
$$
The rotation speed of the earth at the given latitude is just about $390$ m/s, so to account for the plane's speed we can multiply $\theta$ by $0.9$, which multiplies the second row of the table by $0.81$.
Added: The graph shows the altitude of the earth's shadow in this model (neglecting the ground speed of the plane) as a function of latitude (in $2^{\circ}$ increments) and time after sea level sunset, with $\phi = 32.5^{\circ}$ in light blue.
The world being what it is, I should add that while every effort has been made to provide accurate qualitative estimates, this answer is provided for amusement only, and is not intended to be used as a recommendation for actual flight. :)
