We can get a quick upper bound as follows: let $\|x\|$ denote the norm (Euclidean length) of a vector, and denote
$$
\|(x_1,\dots,x_n)\|_\infty = \max_{i=1,\dots,n}|x_n|
$$
It is well known that
$$
\max_{x \neq 0} \frac{\|Ax\|_\infty}{\|x\|_\infty} = \max_{i=1,\dots,n}\sum_{j=1}^n |a_{ij}|
$$
Thus, for a stochastic matrix $A$, we have
$$
\sigma_1(A) = \max_{x \neq 0} \frac{\|Ax\|}{\|x\|} \leq \max_{x \neq 0} \frac{\sqrt{n} \|Ax\|_\infty}{\frac 1{\sqrt{n}}\|x\|_\infty} = n \max_{i=1,\dots,n}\sum_{j=1}^n |a_{ij}| = n
$$
I'm not sure if this bound is tight.
We can see that it is possible to get a maximum singular value of at least $\sqrt{n}$. In particular, it suffices to consider $A = \mathbf 1 e_1^T$
where $\mathbf 1$ is the column-vector of $1$s and $e_1$ is the standard basis vector $(1,0,\dots,0)^T$.