The information you have does not determine the dominant eigenvector.
Let $G$ be the graph with vertex set $\{0,1,\ldots,7\}$ and adjacency matrix
$$
\left(\begin{array}{rrrrrrrr}
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right)
$$
Construct a second graph $H_2$ by joining a new vertex to the vertex 2, and a third
graph $H5$ by joining a new vertex to vertex 5. Then
$$
(A(H_2)^k e)_2 = (A(H_5)^k e)_5
$$
for all $k$. (For $k=0,\ldots,8$ the actual numbers are
$$
1,\\ 4,\\ 8,\\ 25,\\ 57,\\ 163,\\ 392,\\ 1073,\\ 2656)
$$
The Perron vectors are
$$
(1, 1, 1.579071, 1.460275, 1.019079, 1.168003, 0.5330099, 0.206667, 0.612263)
$$
for $H_2$ and, for $H_5$,
$$
(1, 1, 1.579071, 2.0725388, 1.631342, 2.134811, 0.974205, 0.377735, 0.827744)
$$
If you want positive matrices, take the sixth powers of $A(H_2)$ and $A(H_5)$. The relevant
property of $G$ is that the graphs $G\setminus2$ and $G\setminus5$ are cospectral, and their
complements are cospectral too.
All computations carried out in sage.
In a sense the problem is that you are getting a bit of information about each eigenspace,
whereas you want detailed information about a particular eigenspace.
EDIT: this answer was originally written in response to an older version of the question, which merely asked for upper bounds on the largest eigenvalue rather than an algorithm.
Take your matrix to have all entries equal to $1$ to get a matrix which has $n$ as an eigenvalue.
Let $A$ be an $n\times n$ matrix with $0$-$1$ entries. Let $x=(x_1,\dots, x_n)^\top$ and calculate
$$ \begin{aligned}
\sum_{i=1}^n \left\vert \sum_{j=1}^n A_{ij} x_j\right\vert^2
& \leq
n \left( \sum_{j=1}^n \vert x_j\vert \right)^2 \quad\hbox{(triangle inequality)} \\
& \leq
n^2 \sum_{j=1}^n \vert x_j\vert^2 \quad\hbox{(Cauchy-Schwarz)} \\
\end{aligned}
$$
to see that $\Vert Ax \Vert^2 \leq n^2 \Vert x\Vert^2$. So the norm (= largest singular value), and hence the largest eigenvalue, of $A$ is at most $n$. The example I gave at the start shows this is sharp.
The moral is that unless we have more information about this 0-1 matrix, very little informative can be said regarding upper bounds on the largest eigenvalue.
Best Answer
Yes, it is true that the largest eigenvalue is bounded by the largest absolute row sum or column sum. You can check Gershgorin circle theorem. Actually, all the eigenvalues lie in the union of all Gershorin circles.