### Assuming GCH

Here is my original argument, which implicitly assumed $|2^\omega|<|2^{\omega_1}|$ and used an infinite base field.

Take the base field to be $\mathbb Q,$ set $\lambda=2^\omega=\mathfrak c,$ and $X_{\alpha}=\ell^\infty(\alpha)$: the bounded functions $\alpha\to\mathbb R.$ 
(If you prefer, you could use simple functions instead of all bounded functions.) The transition map $X_\alpha\to X_\beta$ is the restriction to $\beta.$

For each $\alpha>0,$ the space $X_\alpha$ has $\mathbb Q$-dimension $\mathfrak c.$ If $\dim X_0$ matters, take the example $X'_\alpha=X_{1+\alpha}$ instead.

For $\alpha<\beta<\omega_1$ the transition map has a kernel of $\mathbb R$-dimension at least one, so $\mathbb Q$-dimension $\mathfrak c.$

The limit $\varprojlim_{\delta<\gamma} X_\delta$ at countable limit ordinals $\gamma$ is the set of functions $\gamma\to\mathbb R$ that are bounded when restricted to $\delta$ for any $\delta<\gamma.$ This set contains $X_\gamma$ as a proper $\mathbb R$-subspace, so the $\mathbb Q$-dimension of $(\varprojlim_{\delta<\gamma} X_\delta)/X_\gamma$ is $\mathfrak c.$

The big limit $\varprojlim_{\alpha < \omega_1} X_\alpha$ is just $\ell^\infty(\omega_1)$ because any unbounded function on $\omega_1$ would be unbounded on some $\alpha<\omega_1.$ The dimension of this space is its cardinality, $2^{\omega_1}.$

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### Not assuming GCH


Let $k$ be a field of cardinality at most $\aleph_0.$
Take $\lambda=\beth_{\omega_1}$ - I need $\lambda^{\aleph_0}=\lambda$ and $\lambda<\lambda^{\aleph_1}.$

Let $K$ be the transcendental extension $k(X_{i,j}:i\in\omega, j\in\lambda).$ So $\dim_k K=|K|=\lambda.$
For any function $f$ let $\|f\|$ denote the least $n\in\omega$ such that the image of $f$ lies in $k(X_{i,j}:i<n, j\in \lambda),$ or $\infty$ if no such $n$ exists. Let $X_\alpha$ be the set of functions $\alpha\to K$ such that $\|f\|<\infty.$ The transition map $X_\alpha\to X_\beta$ is the restriction to $\beta.$
(This bound is a similar idea to that in Tim Campion's answer https://mathoverflow.net/a/376790/164965.)

For each $0<\alpha<\omega_1,$ the space $X_\alpha$ is a $K$-vector space of $K$-dimension at least $2^{\aleph_0}\geq\lambda$ and at most $|\lambda^{\alpha}|=\lambda,$ so its $k$-dimension is $\lambda.$

For $\alpha<\beta<\omega_1$ the transition map has a kernel of $K$-dimension at least one, so $k$-dimension $\lambda.$

The limit $\varprojlim_{\delta<\gamma} X_\delta$ at countable limit ordinals $\gamma$ is the set of functions $f:\gamma\to K$ with $\|f|_\delta\|<\infty$ for each $\delta<\gamma.$ This set contains $X_\gamma$ as a proper $K$-subspace, so the $k$-dimension of $(\varprojlim_{\delta<\gamma} X_\delta)/X_\gamma$ is $\lambda.$

The limit $\varprojlim_{\alpha < \omega_1} X_\alpha$ is $X_{\omega_1}$: its elements $f:\omega_1\to K$ satisfy $\|f\|=\sup_{\alpha<\omega_1} \|f|_\alpha\|<\infty$ because any non-decreasing $\omega_1\to\omega$ is bounded. The $k$-dimension of $X_{\omega_1}\subset K^{\omega_1}$ is upper bounded by $\lambda^{\aleph_1},$ and the $k$-dimension of $\{f:\omega_1\to k(X_{0,j}:j\in\lambda)\}$ is at least $\lambda^{\aleph_1}.$
So $\dim X_{\omega_1}=\lambda^{\aleph_1}.$