To amplify on Brendan's answer:

You are dealing with a matrix of zero mean iid's plus a rank one perturbation:
$M=W+ A$ where $A=c11^T$ and $c=EY$. The spectral norm of $W$ is asymptotically
$2\sqrt{n} \sigma$ where $\sigma^2=\mbox{ Var}(Y)$, see
Geman's 1980 paper (at least when $Y$ has good moment growth). However, the rank one perturbation has norm $cn$, as can be seen by multiplication with the vector
$n^{-1/2} (1,\ldots,1)$. So the norm of $M$ is about $cn$ with error at most
$2\sqrt{n}\sigma$, with high probability.

If $Y$ has heavy tail, everything changes - the norm may be dominated by the maximal entry of $M$. This is the case if the entries are for example $\alpha$-stable with $\alpha<2$; In particular, for one sided Cauchy the norm could be as large as $n^2$.

Edit: The answer above deals with the spectral norm, not the spectral radius. However, in the case of good moment bounds of the entries, it also applies to the spectral radius. This is due to separation of eigenvalues of the matrix $A$, and in particular the fact that the top eigenvalue of $A$
(which equals $cn$) is separated from other eigenvalues of $A$ $(=0)$ by more than twice the norm of $W$. See e.g. Theorem 2 of Schonhage's paper http://www.sciencedirect.com/science/article/pii/002437957990154X
(with k=m=n) and references there to earlier work.