Of course, Igor's answer points the way to working out the answer the OP wanted, but it may not be clear, even after you have got the eigenvalues, what the corresponding eigenfunctions are, or that they have a simple geometric interpretation analogous to the one for the sphere, as the OP asks.

The nice way to describe the first eigenfunctions on the $n$-sphere with its standard metric is to start with the standard isometric embedding $X:S^n\to\mathbb{R}^{n+1}$ and then, for each vector $v\in\mathbb{R}^{n+1}$, one defines the function $f_v:S^n\to\mathbb{R}$ by
$$
f_v(p) = v\cdot X(p).
$$
Every eigenfunction for the first nonzero eigenvalue of the Laplacian is of this form for a unique $v$.

Similarly, for $\mathbb{CP}^n$, there is a natural embedding $X:\mathbb{CP}^n\to\mathsf{H}_{n+1}$, where $\mathsf{H}_{n+1}$ is the (real) vector space of $(n{+}1)$-by-$(n{+}1)$ Hermitian symmetric matrices, given by
$$
X\bigl([v]\bigr) = \frac{v\ v^\ast}{|v|^2} = \frac{v\ v^\ast}{(v^\ast v)},
$$
where $v$ is any nonzero vector in $\mathbb{C}^{n+1}$ (thought of as column vectors of height $(n{+}1)$ and $v^\ast$ is its conjugate transpose, and $[v]\in\mathbb{CP}^n$ is the line in $\mathbb{C}^{n+1}$ that is spanned by $v$. (With the right scaling and inner product on $\mathsf{H}_{n+1}$, this $X$ is an isometric embedding of $\mathbb{CP}^n$ endowed with the Fubini-Study metric. In some sense, it is the simplest such. Note that it is equivariant with respect to the natural actions of $\mathrm{SU}(n{+}1)$ on the domain and range.)

Now let $w\in\mathsf{H}_{n+1}$ be traceless and define
$$
f_w\bigl([v])\bigr) = \mathrm{tr}\bigl(\ w X\bigl([v]\bigr)\ \bigr).
$$
Then every eigenfunction of lowest nontrivial eigenvalue of the Laplacian is of the form $f_w$ for some traceless $w\in\mathsf{H}_{n+1}$. Thus, as in the case of the $n$-sphere, the lowest eigenfunctions are essentially the components of the 'natural' isometric embedding of $\mathbb{CP}^n$. (The reason for the 'essentially' qualifier is that, as defined, $X$ actually maps $\mathbb{CP}^n$ into the hyperplane in $\mathsf{H}_{n+1}$ that consists of the matrices of trace $1$, and one really ought to subtract the term $\tfrac1{n+1}\ I_{n+1}$ from $X$ so that it maps into the hyperplane of traceless matrices. Then the statement is literally true. Those who are familiar with symplectic geometry will recognize that if one then multiplies by $i$, so that the map goes into the traceless *skew*-Hermitian matrices, then the result is the canonical embedding of $\mathbb{CP}^n$ as a (co)-adjoint orbit of $\mathrm{SU}(n{+}1)$.)

To get something like the formula in terms of distance that held for the $n$-sphere, you just need to observe that as $[v]$ varies, the traceless elements
$$
W\bigl([v]\bigr) = X\bigl([v]\bigr) - \tfrac{1}{n+1} I_{n+1}
$$
span the space of traceless Hermitian $(n{+}1)$-by-$(n{+}1)$ matrices, so this means that this eigenspace is spanned by the functions of the form
$$
h_{[w]}\bigl([v]\bigr) = \frac{\bigl|\langle w,v\rangle\bigr|^2}{|w|^2|v|^2} - \frac{1}{n+1}
$$
as $[w]$ varies over $\mathbb{CP}^n$. Obviously, the function $h_{[w]}$ can be expressed as a function of the distance from $[w]$ in the Fubini-Study metric, even though the generic linear combinations of such eigenfunctions aren't usually expressible in terms of distance from a point.