Mathematical Solution

Given two distinct real numbers \( \ell \) and \( n \) such that \( \ell, n > 1 \), the arithmetic mean (A.M.) is:

\[ m = \frac{\ell + n}{2} \]

The geometric means \( G_1, G_2, G_3 \) form a geometric sequence:

\[ \ell, G_1, G_2, G_3, n \]

Let the common ratio be \( r \), then:

\[ G_1 = \ell r, \quad G_2 = \ell r^2, \quad G_3 = \ell r^3, \quad n = \ell r^4 \]

Solving for \( r \):

\[ r = \sqrt[4]{\frac{n}{\ell}} \]

Computing the Required Expression

\[ G_1^4 + 2G_2^4 + G_3^4 \]

\[ \ell^4 r^4 + 2\ell^4 r^8 + \ell^4 r^{12} \]

Factoring \( \ell^4 \):

\[ \ell^4 (r^4 + 2r^8 + r^{12}) \]

Since \( r^4 = \frac{n}{\ell} \), we substitute:

\[ \ell^4 \left( \frac{n}{\ell} + 2 \frac{n^2}{\ell^2} + \frac{n^3}{\ell^3} \right) \]

Multiplying:

\[ \ell^3 n + 2\ell^2 n^2 + \ell n^3 \]

Expressing in Terms of \( m \)

\[ m^2 = \left( \frac{\ell + n}{2} \right)^2 = \frac{\ell^2 + 2\ell n + n^2}{4} \]

Multiplying by \( 4 \ell n \):

\[ 4 \ell n m^2 = \ell^3 n + 2\ell^2 n^2 + \ell n^3 \]

Final Answer

\[ \mathbf{4 \ell m^2 n} \]