As is discussed in this answer, the connected sum of two copies of $\mathbb{RP}^3$ can be realised as the compact affine grassmannian of $2$-planes in $\mathbb{R}^3$, denoted by $\operatorname{cGraff}_2(\mathbb{R}^3)$, and there is a fiber bundle
$$S^1 \to \operatorname{cGraff}_2(\mathbb{R}^3) \to \mathbb{RP}^2.$$
Is there a fiber bundle whose total space is the connected sum of three copies of $\mathbb{RP}^3$?
Let $3\mathbb{RP}^3$ denote the connected sum of three copies of $\mathbb{RP}^3$. If $F \to 3\mathbb{RP}^3 \xrightarrow{\pi} B$ is a fiber bundle with manifold fiber and base, then $F$ and $B$ are closed (compact and without boundary), and $B$ is connected. Note that $\dim F, \dim B \in \{0, 1, 2, 3\}$ and $\dim F + \dim B = \dim 3\mathbb{RP}^3 = 3$.
$\bullet$ If $\dim F = 3$ and $\dim B = 0$, then we have only one possibility: $3\mathbb{RP}^3 \to 3\mathbb{RP}^3 \to \ast$ where $\ast$ denotes a singleton.
$\bullet$ If $\dim F = 2$ and $\dim B = 1$, then we have a fiber bundle of the form $F \to 3\mathbb{RP}^3 \to S^1$. Note that a part of the associated long exact sequence in homotopy is
$$\dots \to \pi_1(3\mathbb{RP}^3) \to \pi_1(S^1) \to \pi_0(F) \to \dots$$
Since $\pi_1(3\mathbb{RP}^3) \cong \mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ has a generating set which are torsion elements and $\pi_1(S^1) \cong \mathbb{Z}$ is torsion-free, the map $\pi_1(3\mathbb{RP}^3) \to \pi_1(S^1)$ is trivial, so $\pi_1(S^1)$ injects into $\pi_0(F)$. But this would imply that $F$ has infinitely many components, contradicting the compactness of $F$.
$\bullet$ If $\dim F = 1$ and $\dim B = 2$, then we have a fiber bundle $F \to 3\mathbb{RP}^3 \to B$ where $F$ is a disjoint union of finitely many copies of $S^1$, and $B$ is a closed surface. Again, from the long exact sequence in homotopy we have
$$\dots \to \pi_1(3\mathbb{RP}^3) \to \pi_1(B) \to \pi_0(F) \to \dots$$
If $B \neq S^2$ or $\mathbb{RP}^2$, then $\pi_1(B)$ is torsion-free and infinite, so by the same argument as in the previous case, $F$ would have to have infinitely many components which is impossible.
If $B = S^2$, then from the long exact sequence in homotopy we have
$$\dots \to \pi_1(F) \to \pi_1(3\mathbb{RP}^3) \to \pi_1(S^2) \to \dots$$
Since $S^2$ is simply connected, we see that $\pi_1(F) \cong \mathbb{Z}$ surjects onto $\pi_1(3\mathbb{RP}^3) \cong \mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ which is impossible as the only infinite quotient of $\mathbb{Z}$ is $\mathbb{Z}$ itself and $\mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2\not\cong\mathbb{Z}$.
If $B = \mathbb{RP}^2$, we instead obtain
$$\dots \to \pi_1(F) \to \pi_1(3\mathbb{RP}^3) \xrightarrow{\alpha} \pi_1(\mathbb{RP}^2) \to \dots$$
where now $\pi_1(\mathbb{RP}^2) \cong \mathbb{Z}_2 \not\cong 0$. If the map $\alpha$ is trivial, then the same argument as in the $S^2$ case applies. Suppose then that $\alpha$ is non-trivial (and hence surjective). Then we see that $\pi_1(F) \cong \mathbb{Z}$ surjects onto the kernel of $\alpha$. Since $\pi_1(3\mathbb{RP}^3) \cong \mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ is infinite and $\mathbb{Z}_2$ is finite, we see that $\ker\alpha$ is infinite. Again, the only infinite quotient of $\mathbb{Z}$ is $\mathbb{Z}$ itself, so we must have $\mathbb{Z} \cong \ker\alpha$. Let $a, b, c$ denote the standard generators of $\mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$. Since these elements are torsion, they do not belong to the kernel of $\alpha$ as $\ker\alpha\cong\mathbb{Z}$ is torsion-free. It follows that $ab$ and $ac$ do belong to the kernel which is a contradiction as one must be a power of the other since $\ker\alpha$ is infinite cyclic, but the only relations in $\mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ are $a^2 = b^2 = c^2 = 1$.
$\bullet$ If $\dim F = 0$ and $\dim B = 3$, then $3\mathbb{RP}^3$ is a finite covering space of a connected closed three-manifold. Of course, there is the trivial covering $\ast \to 3\mathbb{RP}^3 \to 3\mathbb{RP}^3$, but $3\mathbb{RP}^3$ can also be realised as the total space of a non-trivial covering. Let $L(4; 1, 1)$ be the lens space $S^3/\mathbb{Z}_4$ where $\mathbb{Z}_4$ acts by $(z_1, z_2) \mapsto (e^{\pi i/2}z_1, e^{\pi i/2}z_2)$. This is double covered by $\mathbb{RP}^3$, see here. Therefore, $\mathbb{RP}^3\# 2\mathbb{RP}^3 = 3\mathbb{RP}^3$ double covers $L(4; 1, 1)\#\mathbb{RP}^3$ by the useful trick mentioned in this answer. Similarly, since $S^3$ is a triple cover of $L(3; 1, 1)$, we see that $S^3\# 3\mathbb{RP}^3 = 3\mathbb{RP}^3$ is a triple cover of $L(3; 1, 1)\#\mathbb{RP}^3$.
Conclusion: $3\mathbb{RP}^3$ cannot be realised as the total space of a fiber bundle with positive-dimensional manifold fiber and base, but it can be realised as the total space of a non-trivial covering map.