My reasoning is a bit long. In spite of this I had to omit some details.
Let $G$ be a planar graph with $13$ vertices and the degree of each vertex of $G$ be at least $5$.
We can assume that each face of $G$ is a triangle and then each edge of $G$ lies exactly in two faces.
We will prove that the existence of such a graph leads to a contradiction.
Lemma 1. The graph $G$ has exactly one vertex of degree $6$ and $12$ vertices of degree $5$.
We assume that the vertices of $V(G)$ lie on some plane, and the edges of $G$ are some curves connecting the vertices, and any two edges have no common points in the plane except vertices.
Lemma 2. If $T=x_1x_2x_3$ is a triangle in $G$ (that is, $x_1,x_2,x_3\in V(G)$ and $x_1x_2,x_2x_3,x_3x_1\in E(G)$), then either
no vertices of $G$ lie inside $T$, or all vertices of $G$ lie inside $T$ (except
vertices $x_i$ of course).
In other words, each triangle of graph $G$ is an face.
Proof.
Let $k$ vertices of graph $G$ lie inside $T$ and $k>0$.
Obviously, $k\geq3$.
We consider only the case when $k=5$. Leave the cases $k=3,4$ as an easy exercise.
If $k>5$, then $l=10-k\leq4$ and there are exactly $l\leq4$ vertices outside $T$.
Hence the same arguments show that it should be $l=0$ or $k=10$.
So let $k=5$.
Let $v_1,v_2,v_3\in V(G)$ and the faces $x_1x_2v_3$, $x_2x_3v_1$,
and $x_1x_3v_2$ lie inside triangle $T$.
It is easy to see that $v_1,v_2,v_3$ are pairwise different
(if for example $v_1=v_2$, then $T$ splits into three triangles
and no more than $4$ vertices lie inside one of them).
It follows that in the induced graph $H$ with set of vertices
$\{x_1,x_2,x_3,v_1,v_2,v_3,v_4,v_5\}$,
where $v_3,v_4$ lie inside $T$, we have $\deg_H(x_i\geq4)$, $i=1,2,3$.
Then by the handshake lemma the following holds
$$
2|E(H)|\geq3\cdot4+5\cdot5=37\ \Rightarrow\ E(H)\geq19.
$$
On the other hand
$$
|E(H)|\leq3\cdot8-6=18.
$$
We get a contradiction.
Lemma 3. If $Q=x_1x_2x_3x_4$ is a quadrilateral in $G$ ( i.e., $Q$ is a cycle in $G$ of length $4$ and $x_1x_2,x_2x_3,x_3x_4,x_4x_1\in E(G)$),
then either no vertices lie inside $Q$, or all vertices lie
inside $Q$ (except vertices $x_i$).
Proof.
Let $k$ vertices of graph $G$ lie inside $Q$ and $k>0$.
It is sufficient to consider the cases $k=1,2,3,4$.
For $k>4$ we have $l=9-k\leq4$ and we can go to the outer points of $Q$.
Then it is easy to see that $k\geq3$.
Of the two remaining cases we will consider only the case $k=4$.
As in the previous lemma, let $x_1x_2v_1$, $x_2x_3v_2$, $x_3x_4v_3$, $x_4x_1v_4$ be faces lying inside $Q$.
It is easy to see that
all $v_i$ are pairwise different,
each $v_i$ is adjacent only to two vertices of $Q$,
the induced graph $F$ with set of vertices $v_i$ is a cycle.
It follows that $\deg(v_i)<5$. Contradiction.
Lemma 4. Let $v\in V(G)$, $\deg(v)=6$ and $u_i$ ($1\leq i\leq 6$) be all neighbors of the vertex $v$. Then the induced graph $H$ with
set of vertices $u_i$ is a cycle.
Proof.
This follows from Lemmas 2 and 3.
Then we reason like this. Let $v\in V(G)$, $\deg(v)=6$.
Let $u_i$ ($1\leq i\leq 6$) be all neighbors of the vertex $v$.
Let $u_iu_{i+1}w_i$ ($i=1,\ldots,6$, $u_7=u_1$) be faces lying inside $H$ (see figure).
The vertices $w_i$ are pairwise different and distinct from vertices $u_i$, $v$.
If vertex $w_1$ is adjacent to vertex $w_3$, then there are vertices inside and outside the quadrilateral
$w_1w_3u_3u_2$ (contrary to Lemma 3).
If vertex $w_1$ is adjacent to vertex $w_4$, then vertex $w_2$ is adjacent to vertex $w_4$
(otherwise $\deg(w_2)<5$). We come to the case discussed in 2).
This concludes our proof.
Best Answer
One possible construction is a silly generalization of the icosahedron.
Start with the skeleton graph of an arbitrarily large antiprism: two $k$-sided polygons connected by a zigzagging strip of $2k$ triangles. Here, every vertex has degree $4$.
Then, add two more vertices, one in the middle of each length-$k$ face. The old vertices now have degree $5$, and the two new vertices have degree $k$. So, for all $k \ge 5$, this gives us a $(2k+2)$-vertex planar graph with minimum degree $5$. (When $k=5$, that graph is the skeleton graph of the icosahedron.)
A new vertex is at distance $1$ from each vertex of the length-$k$ face it was placed in, at distance $2$ from each vertex of the opposite length-$k$ face, and at distance $3$ from the other new vertex.
An old vertex is at distance at most $2$ from each other vertex of the same length-$k$ face, by going through the new vertex they're both adjacent to. So an old vertex is at distance at most $3$ from each other vertex of the opposite length-$k$ face (we start by going to a neighbor on the opposite face, and then finish as above). Therefore the diameter is $3$.
Here is a diagram of the $k=10$ case. It is not drawn in a planar fashion, because I see no clean way to do that, but you can imagine moving one of the central vertices (the one to the left, specifically) to the outside and drawing lots of curved edges.
If we want to generalize to an odd number of vertices, we can do that. In one of the length-$k$ faces, instead add two vertices with an edge between them, each adjacent to half of the length-$k$ face. (We can do slightly more than half, really; e.g. when $k=6$, each of the two vertices can be adjacent to $4$ out of $6$ vertices on the length-$6$ face.) This graph has $2k+3$ vertices, still has diameter $3$, and for $k \ge 6$ it has minimum degree $5$.
The $(2k+2)$-vertex construction gives us all even $n \ge 12$, and the $(2k+3)$-vertex construction gives us all odd $n\ge 15$. Therefore an $n$-vertex graph is achievable for all $n \ge 12$ except for $n=13$. On the other hand, a planar graph with minimum degree $5$ must have at least $12$ vertices of degree exactly $5$, so $n<12$ is impossible, and Plantri tells me that there are no $13$-vertex planar graphs with minimum degree $5$ at all! So the construction above covers all the values of $n$ we could have hoped for.
While we're at it, let's check the connectivity.
The antiprism graph is $4$-connected. To see this, suppose only $3$ vertices are deleted from the antiprism. Then one of the two length-$k$ faces only loses a single vertex, so it is still connected. Every vertex in the other face is adjacent to two vertices of the first face, at least one of which was not deleted, so it is also part of that connected component. Therefore deleting $3$ vertices does not disconnect the antiprism.
Both the odd and the even constructions above are $5$-connected. To see this, suppose only $4$ vertices are deleted from the graph. There are two cases: