In general it could happen that the Albanese variety does not admit a principal polarization at all. For instance the Albanese variety of an abelian variety is the Abelian variety itself. So choose $X$ to be some abelian variety that has no principal polarization and you will get an example.

On the other hand it can happen that the Albanese variety is principally polarized. For instance you can take the Albanese of the $n$-th symmetric product of a curve. It is equal to the Jacobian of the curve and so admits a principal polarization. Or if you want to be fancier you can take a hyperplane section in the symmetric product of a curve. It will also have the Jacobian of the curve as its Albanese variety.

Another useful comment is that the Albanese of $X$ is the dual of $Pic^{0}(X)$ and so $Alb(X)$ admits a principal polarization if and only if $Pic^{0}(X)$ does. If you fix an ample line bundle $L$ on an $n$-dimensional complex projective variety $X$, then $L$ induces a natural polarization on $Pic^{0}(X)$: the universal cover of $Pic^{0}(X)$ is naturally identified with $H^{1}(X,O_{X}) = H^{1,0}(X)$, the integral $(1,1)$ form $c_{1}(L)$ then induces a Hermitian pairing on $H^{1,0}(X)$ by the formula
$$
h(\alpha,\beta) := -2i \int_{X} \alpha\wedge \bar{\beta} \wedge c_{1}(L)^{\wedge (n-1)}.
$$
This $h$ defines a polarization on $Pic^{0}(X)$. The construction of $h$ is purely cohomological and so it is straightforward to check if it defines a principal polarization by computing the divisors of this polarization.