The following partial answer applies to matrices $A \ \in \ M^{n \times n}$ of the question, if $A$ is also symmetric (semidefinite in the narrow sense), but the principles will be useful to understand the general case, I suppose.
The first version contained a stupid mistake, which I corrected after I found my first solution to be inconsistent with the solution to the second problem linked to in the question.
As our matrix $A$ in question is symmetric, it can be written as
$$ A \ = \ \lambda_1 v_1v_1^T + \lambda_2 v_2v_2^T + \ldots + \lambda_k v_kv_k^T \, ,$$
where the $\lambda_i$ are its non-vanishing singular values and eigenvalues in this special case $(k \ \leq \ n)$, which we have arranged in decreasing order, so $\lambda_1$ is a Perron root of $A$, and the $v_i$ are corresponding orthonormal eigenvectors. We allow a multiple Perron root, hence there is no irreducibility assumption. We can take $v_1$ to have non-negative entries though, which we shall do.
The Moore-Penrose inverse is then given as
$$ A^\dagger \ = \ \frac{1}{\lambda_1} v_1v_1^T + \frac{1}{\lambda_2} v_2v_2^T + \ldots + \frac{1}{\lambda_k} v_kv_k^T \, .$$
The key observation following an elementary and simple computation is
$$ e^T(xx^T)e \ = \ (c_x)^2 \, ,$$
where $x$ is an arbitrary column vector and $c_x$ is its component sum, so the result is non-negative, if $x$ contains real numbers as entries.
As the entries of $v_1$ are non-negative and $v_1$ has unit length, its component sum is at least $1$ and at most $\sqrt{n}$, which occurs if and only if all components are equal. But as the $v_i$ form an orthonormal system, all other $v_i$ must have vanishing component sums in this case, as they are orthogonal to $e$ in particular. Therefore we have
$$ e^T A^\dagger e \ = \ \frac{n}{\lambda_1}$$
in this case, while we have
$$ e^T A^\dagger e \ = \ \sum_{i=1}^k \frac{(c_{v_i})^2}{\lambda_i}$$
and therefore
$$ \frac{1}{\lambda_1} \ \leq \ e^T A^\dagger e \ \leq n \sum_{i=1}^k \frac{1}{\lambda_i}$$
in general.
In particular, we have $$e^T A^\dagger e \ = \ n$$ if $A$ is doubly stochastic.