This is a very interesting question, and (remarkably) we can decide it entirely using only relatively basic tools from the algebraic topologist's tool case. To be precise, we're going to prove the following:
Proposition: A map $\newcommand{\RP}{\mathbb{R}\mathrm{P}}f\colon \RP^n \to M$ of degree 1 where $n$ is odd and $M$ a closed orientable $n$-manifold exists if and only if $M$ is either $S^n$ or homotopy equivalent to $\RP^n$.
To prove this, first note that since the quotient/covering map $S^n \to \RP^n$ is of degree 2, any such $f$ gives rise to a degree 2 map $S^n \to M$ just by precomposition, so by a standard application of Poincaré duality all the homology groups $H_k(M)$ for $0 < k < n$ must be 2-torsion (see this question for more details). If $H_k(M) = 0$ for all $0 < k < n$, then $M$ is an integral homology sphere, but then $M$ must already be simply connected (see KFJ2611's comment about $\pi_1(M)$ either being 0 or $\mathbb{Z} / 2$) and therefore homeomorphic to $S^n$ (by Whitehead's theorem applied to a degree 1 map $M \to S^n$ and the generalised Poincaré conjecture¹).
Otherwise, we will show that $f_*\colon H_*(\RP^n) \to H_*(M)$ is already an isomorphism.
To this end, note first of all that $f_*\colon H_*(\RP^n) \to H_*(M)$ is surjective since $f$ is of degree 1 (this is a consequence of the projection formula $\alpha \frown f_*(b) = f_*(f^*(\alpha) \frown b)$ for all $\alpha \in H^*(M)$, $b \in H_*(\RP^n)$ applied to $b$ the fundamental class of $\RP^n$ and Poincaré duality, though let me know if you want more details). Thus, $H_k(M)$ is 0 if $k$ is even and either $\newcommand{Z}{\mathbb{Z}}\newcommand{\F}{\mathbb{F}} \Z / 2$ or 0 if $k$ is odd in the range $0 < k < n$. Passing to co/homology with coefficients in $\F_2$, we obtain a map from a graded $\F_2$-algebra $H^*(M; \F_2)$ which is levelwise of dimension at most 1 to $H^*(\RP^n; \F_2) \cong \F_2[x]$ with $x$ the unique class in degree 1.
Let now $0 < k < n$ be the lowest degree in which $H^*(M; \F_2) \neq 0$. If $k = 1$, then since $f^*(y_1^k) = f^*(y_1)^k = x^k$ for $y_1 \in H^1(M; \F_2)$ the generator we see that $f^*$ is an isomorphism, hence $f_*$ is an isomorphism both with $\F_2$ and integral coefficients and we are done². Otherwise, we note that by the universal coefficient theorem we must have $H^{k + 1}(M; \F_2) \neq 0$ as well, so by Poincaré duality $H^{n - k}(M; \F_2)$ and $H^{n - k - 1}(M; \F_2)$, too, must be non-zero, whence $y_k y_{n - k - 1}$ is a nonzero element in degree $n - 1$ (again since $f^*(y_k y_{n - k - 1}) = f^*(y_k) f^*(y_{n - k - 1}) = x^k x^{n - k - 1} = x^{n - 1} \neq 0$) and another round of Poincaré duality then yields that $H^1(M; \F_2) \neq 0$, contradicting our choice of $k$. $\Box$
¹I know I promised "only relatively basic tools from the algebraic topologist's tool case." This is the one "advanced" theorem I need.
²As was pointed out by ShamanR in the comments, it does not directly follow from this that $M$ is homotopy equivalent to $\RP^n$. To conclude, we need the following version of the Whitehead/Hurewicz theorem:
Theorem: Let $g\colon X \to Y$ be a map between CW-complexes. If $g$ induces an isomorphism on $\pi_1$ and moreover a lift $\tilde{g}\colon \tilde{X} \to \tilde{Y}$ of $g$ to the universal covers of $X$ and $Y$ induces an isomorphism on integral homology, then $g$ is a homotopy equivalence.
Proof: By Whitehead's theorem $\tilde{g}_*$ induces an isomorphism on homotopy groups since $\tilde{X}$ and $\tilde{Y}$ are simply connected, and since covering projections induce isomorphisms on $\pi_k$ for all $k > 1$ we see that $g$ must induce an isomorphism on all homotopy groups. $\Box$
Derived Cats was so kind as to fill in the rest of the argument:
Let $\tilde{M}$ be the universal cover of $M$. Since $\pi_1(M) \cong \Z / 2$ it is a double cover, so a lift $\tilde{f}\colon S^n \to \tilde{M}$ of $f$ to universal covers must be a degree 1 map (here we use that $\tilde{M}$ is itself a closed orientable $n$-manifold). By the same argument as in the first paragraph of the proof, this means that $\tilde{M}$ is a simply connected integral homology sphere and therefore homeomorphic to $S^n$. Thus $\tilde{f}_*\colon H_*(S^n) \to H_*(\tilde{M})$ is an isomorphism and we conclude by the theorem after noting that it is applicable since closed manifolds have the homotopy type of a CW-complex.
