(This question, as posted earlier, had not specified that the heptagon must be convex. My answer here addresses the later version of the question, which specifies that it must be convex.)

No such convex heptagon exists. At any point $P$ where 3 or more tiles meet, the mean of the angles at $P$ is at most $2\pi/3$. A heptagon's interior angles sum to $5\pi$ so its mean angle is $5\pi/7>2\pi/3$. Sustaining so high a mean entails points where the interior angles of only 2 tiles meet. At any such point $P$, one of two possibilities occurs.

One possibility is that one of those angles is greater than $\pi$, i.e. it is a reflex angle, which means that the tile containing it is concave.

The other possibility is that both of the angles there are equal to $\pi$. But in that case, $P$ is not a vertex of either of the 2 tiles meeting there. To call $P$ a point where two vertices meet would be to count 2 or more pieces of a single side as separate sides; the "heptagon" would in fact have no more than 6 sides.

Note that it is OK for 3 or more tiles to meet at $P$ and for one of the angles at $P$ to be $\pi$. This would mean that vertices of 2 or more tiles met at $P$ along with some point of another tile which is not a vertex.

(This question, as posted earlier, had not specified that the heptagon must be convex. My answer here addresses the later version of the question, which specifies that it must be convex.)

No such convex heptagonal tile $T$ exists. At any point $P$ where 3 or more tiles meet, the mean of the angles at $P$ is at most $2\pi/3$. $T$'s interior angles sum to $5\pi$ so $T$'s mean interior angle $M(T)=5\pi/7>2\pi/3$.

Consider the "mean angle of a tiling" defined as follows. Pick an arbitrary point $O$ in the tiling. For any radius $r>0$, let $C(r)$ be the circle centre $O$, radius $r$. Consider the set $S$ of all points $P$ where the interior angles of 2 or more tiles meet, where $|OP|<r$. What is the mean $M(r)$ of all these interior angles? It is a weighted average of the respective averages over all points in $S$; the weight for $P$ being proportional to the number of angles meeting at $P$. For large $r$, $M(r)\approx M(T)$. The two means might not be exactly equal, because of tiles that straddle $C$ and have at least one vertex within $C(r)$ and at least one outside $C(r)$. But such tiles can be made an arbitrarily small proportional of the total by taking $r$ large enough. Thus $$\lim_{r\to\infty} M(r)=M(T).$$

As shown above, $M(r)$ cannot be made that high using only points where 3 or more tiles meet. This entails points where the interior angles of only 2 tiles meet. At any such point $P$, one of two possibilities occurs.

One possibility is that one of those angles is greater than $\pi$, i.e. it is a reflex angle, which means that the tile containing it is concave.

The other possibility is that both of the angles there are equal to $\pi$. But in that case, $P$ is not a vertex of either of the 2 tiles meeting there. To call $P$ a point where two vertices meet would be to count 2 or more pieces of a single side as separate sides; the "heptagon" would in fact have no more than 6 sides.

Note that it is OK for 3 or more tiles to meet at $P$ and for one of the angles at $P$ to be $\pi$. This would mean that vertices of 2 or more tiles met at $P$ along with some point of another tile which is not a vertex.