It is unlikely that you will get an answer to your question which relies on off-the-shelf algorithms, such as the ones in Mathematica. For what it is worth, an approximate answer which seems to be very close to the global maximum is:
sol = {c -> -4.84816358599152`, d -> 0.3008466924271099`, e -> -1172.3619400404964`, m -> -357.535888933738`, n -> -1267.447358996882`}
which is such that
Min[Eigenvalues[Q /. sol]]
is
3.55232
As for why this is close to the optimum consider the following identity:
$$
\begin{bmatrix}
a_1 & a_2 \\
a_2 & a_3
\end{bmatrix}
=
\begin{bmatrix}
c \\
d
\end{bmatrix}
\begin{bmatrix}
c &
d
\end{bmatrix}
=
\begin{bmatrix}
c^2 & c d\\
c d & d^2
\end{bmatrix}
$$
A relaxation of this into the matrix inequality
$$
\begin{bmatrix}
a_1 & a_2 \\
a_2 & a_3
\end{bmatrix}
\succeq
\begin{bmatrix}
c \\
d
\end{bmatrix}
\begin{bmatrix}
c &
d
\end{bmatrix}
$$
can be restated as the linear matrix inequality
$$
S(c,d,a1,a2,a3) =
\begin{bmatrix}
a_1 & a_2 & c\\
a_2 & a_3 & d \\
c & d & 1
\end{bmatrix}
\succeq
0.
$$
Now let $Q(c,d,e,m,n,c^2,c d, d^2)$ denote the original matrix $Q$. An upper bound to the maximum of the minimum eigenvalue is given by the solution to the semidefinite program
$$
\max_{c,d,e,m,n,a1,a2,a3} \{ \rho : \quad Q(c,d,e,m,n,a1,a2,a3) - \rho I \succeq 0, \quad S(c,d,a1,a2,a3) \succeq 0 \}.
$$
This is a semidefinite program and it can be solved in Mathematica as follows:
Qr = (Q /. {c^2 -> a1, c d -> a2, d^2 -> a3}) - \[Rho] IdentityMatrix[4];
Sr = {{a1, a2, c}, {a2, a3, d}, {c, d, 1}};
sol = SemidefiniteOptimization[-\[Rho], VectorGreaterEqual[{Qr, 0}, {"SemidefiniteCone", 4}] && VectorGreaterEqual[{Sr, 0}, {"SemidefiniteCone", 3}], Variables[Qr]]
which is such that
\[Rho] /. sol
evaluates to
3.55235
The gap between the relaxation and the solution provided before is of order of $10^{-5}$. Beware not to use any other components of the relaxation solution since they are not feasible to the original problem. Its value is only in certifying the (near) global optimality of the other solution.
