Here is an observation in the positive direction:

Observation: Let $\mathcal C$ be a cocomplete category, and assume that $I \in \mathcal C$ is a strong cogenerator -- that is, $Hom_{\mathcal C}(-,I) : \mathcal C^{op} \to Set$ is faithful and conservative. Let $\mathcal D \subseteq \mathcal C$ be a full subcategory which contains $I$. Then the inclusion $\iota: \mathcal D \to \mathcal C$ preserves colimits.

Proof: Suppose that $D = \varinjlim_j^{\mathcal D} D_j$ is a colimit in $\mathcal D$, and let $\underline D = \varinjlim_j^{\mathcal C} D_j$ be the colimit taken in $\mathcal C$. There is a canonical map $\varepsilon: \underline D \to D$. We have $Hom_{\mathcal C}(\underline D, I) = \varprojlim_j Hom_{\mathcal C}(D_j, I) = \varprojlim_j Hom_{\mathcal D}(D_j, I) = Hom_{\mathcal D}(D,I)$, so that $Hom(\varepsilon, I)$ is an isomorphism by Yoneda. Thus $\varepsilon$ is an isomorphism because $I$ is a strong cogenerator.

Example: Euclidean space $\mathbb R^d$ (for $d \geq 1$) is a strong [1] cogenerator in the category $Tych$ of Tychonoff spaces, which is contained in the full subcategory $Man \subset Tych$ of topological manifolds. Therefore any colimit of topological manifolds is a colimit of Tychonoff spaces.

[1] That it's a cogenerator follows almost by definition, but I'm having trouble tracking down a reference that it's a strong cogenerator. I'm pretty confident that the unit interval $[0,1]$ is a strong cogenerator in $Tych$ from the construction of the Stone-Cech compactification, but I have not worked out anything further. So take what I say with a grain of salt.

Here is an observation in the positive direction:

Observation: Let $\mathcal C$ be a cocomplete category, and assume that $I \in \mathcal C$ is a strong cogenerator -- that is, $Hom_{\mathcal C}(-,I) : \mathcal C^{op} \to Set$ is faithful and conservative. Let $\mathcal D \subseteq \mathcal C$ be a full subcategory which contains $I$. Then the inclusion $\iota: \mathcal D \to \mathcal C$ preserves colimits.

Proof: Suppose that $D = \varinjlim_j^{\mathcal D} D_j$ is a colimit in $\mathcal D$, and let $\underline D = \varinjlim_j^{\mathcal C} D_j$ be the colimit taken in $\mathcal C$. There is a canonical map $\varepsilon: \underline D \to D$. We have $Hom_{\mathcal C}(\underline D, I) = \varprojlim_j Hom_{\mathcal C}(D_j, I) = \varprojlim_j Hom_{\mathcal D}(D_j, I) = Hom_{\mathcal D}(D,I)$, so that $Hom(\varepsilon, I)$ is a bijection. Thus $\varepsilon$ is an isomorphism because $I$ is a strong cogenerator.

Example: Euclidean space $\mathbb R^d$ (for $d \geq 1$) is a strong [1] cogenerator in the category $Tych$ of Tychonoff spaces, which is contained in the full subcategory $Man \subset Tych$ of topological manifolds. Therefore any colimit of topological manifolds is a colimit of Tychonoff spaces.

[1] That it's a cogenerator follows almost by definition. So to see that $\mathbb R$ is a strong cogenerator in $Tych$, let $f : X \to Y$ be a map in $Tych$ which induces a bijection $f^\ast : C^0(Y) \to C^0(X)$; we wish to show that $f$ is a homeomorphism. First, $f$ is injective. For if $x \neq x' \in X$, then we may find a function $\varphi \in C^0(X)$ with $\varphi(x) \neq \varphi(x')$; since $\varphi$ extends along $f$ it follows that $f(x) \neq f(x')$. Next, $f$ is surjective. For if $y \in Y$ is not isolated and not in the image of $f$, then let $\psi \in C^0(Y)$ vanish uniquely at $y$; then $1/(f^\ast \psi) \in C^0(X)$ does not extend along $f$. If $f$ misses an isolated point $y \in Y$, then $f^\ast$ is not injective. Thus $f$ is a continuous bijection. If $f$ is not a homeomorphism, then pick a closed set $C \subseteq X$ whose image $f(C) \subseteq Y$ is not closed. Let $\varphi \in C^0(X)$ vanish uniquely on $C$. Then $\varphi$ does not extend continuously to $Y$.